two adjacent angles of a rhombus as in the ratio 2:3. Eind all the a angles of the rombus​

Answer:

x = 80 , y = 50

Step-by-step explanation:

the angle between the tangent and the radius at the point of contact is 90°

given the angle between the tangent and the chord is 40° , then

angle inside triangle = 90° - 40° = 50°

the triangle has 2 equal radii forming 2 sides thus is isosceles with 2 base angles being congruent, then

y = 50°

the sum of the 3 angles in the triangle = 180° , then

x = 180° - 50° - 50° = 180° - 100° = 80°

Answer:

x = 80°

y = 50°

Step-by-step explanation:

the legs of the inner triangle (tangent to O, and O to point with y angle) are equal because both are the radius of the circle.

that makes the inner triangle an isoceles triangle with both angles on the baseline (tangent to point with y angle) being equal.

the angle of the tangent to the leg "tangent to O" is per definition a right angle (90°). otherwise it would not be a tangent.

one post of the right angle is 40°, so the other part (the triangle inner angle at the tangent point) is then 90-40 = 50°.

since both leg angles must be equal (as described above), y = 50° too.

and as the sum of all angles in a triangle must be 180°, that gives us for x

180 = 50 + 50 + x

x = 180 - 50 - 50 = 80°

Answer:

Step-by-step explanation:

70 = x/4

x = 4 cos 70 = 1.37m

x = 4 cos  80 = 0.69m  

The range is ( 0.69m, 1.37m)

The range of distances from the wall (x) that the foot of the ladder  placed to fall within the safety guidelines is between 0.7052 meters and 1.456 meters.

To determine the range of distances from the wall that the foot of the ladder may be placed to fall within the safety guidelines,  use trigonometry.

Let's assume the distance from the wall to the foot of the ladder is x meters the ladder is 4 meters long.

The angle between the ladder and the ground is given to be between 70° and 80°. Let's consider the extreme cases for each angle:

When the ladder makes an angle of 70° with the ground:

In this case, the angle between the wall and the ladder (θ) will be 90° - 70° = 20°.

When the ladder makes an angle of 80° with the ground:

In this case, the angle between the wall and the ladder (θ) will be 90° - 80° = 10°.

Now, use trigonometry to calculate the range of distances (x) from the wall:

For the first case (θ = 20°):

tan(20°) = Opposite / Adjacent

tan(20°) = x / 4

x = 4 × tan(20°)

For the second case (θ = 10°):

tan(10°) = Opposite / Adjacent

tan(10°) = x / 4

x = 4 × tan(10°)

Now, let's calculate the values:

x ≈ 4 × 0.3640 ≈ 1.456 meters (rounded to three decimal places) - for the first case.

x ≈ 4 × 0.1763 ≈ 0.7052 meters (rounded to four decimal places) - for the second case.

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Answer:

FIRST

To find the volume of rectangular prism is

Volume = Length x Width x Height

Volume = 20cm x 10cm x 13cm

= 40 x 13

= 520cm cubic or cube

520 is the volume of the rectangular prism

the you divide volume of rectangular prism and volume of solid ball.

Correct me if i am wrong guys.

Answer:

It will take approximately take 17 balls to overflow the container

Step-by-step explanation:

Volume of the rectangle = 20x13x10 = 2600[tex]cm^{3}[/tex]

Volume of the water = 20x10x11 = 2200[tex]cm^{2}[/tex]

Amount of empty space = 2600-2200 = 400[tex]cm^{3[/tex]

Solid ball volume = 23[tex]cm^{3}[/tex] each

To find how many balls can overflow the container = 400/23 = 17.39

Percent of the original volume that is removed is; 18%

Formula for volume of a box is;

V = lbh

where;

l is length

b is breadth

h is height

Thus;

V_original = 15 * 10 * 8

V_original = 1200 cm³

Volume for each cube removed = 3 * 3 * 3 = 27 cm³

Since there are 8 corners on the box, then 8 cubes are removed.

So the total volume removed is; 8 * 27 = 216

Thus;

Percent of the original volume that is removed is;

216/1200 * 100% = 18%

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From the proof of modular congruence below, it has been shown that;

41 ≡ 21 (mod 3).

We want to use the definition of modular congruence to prove that;

41 is congruent to 21 (mod 3) i.e if a ≡ b (mod m) then b ≡ a (mod m).

We are trying to prove that modular congruence mod 3 is a symmetric relation on the integers.

First, if we recall the definition of modular congruence:

For integers a, b and positive integer m,  

a ≡ b (mod m) if and only if m|a–b

Suppose 41 ≡ 21 (mod 3).

Then, by definition, 3|41–21, so there is an integer k such that 41 – 21 = 3k.

Thus;

–(41 – 21) = –3k

So

21 – 41 = 3(–k)

This shows that 3|21 – 41.

Thus;

21 ≡ 41 (mod 3) and the proof is complete

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[tex]2sin {}^{2} (x) + sin(2x) = 2 \\ 2sin {}^{2} (x) + 2sin(x)cos(x) = 2 \\ sin {}^{2} (x) + sin(x)cos(x) - 1 = 0 \\1 - cos {}^{2} (x) + sin(x)cos(x) - 1 = 0 \\ - cos {}^{2} (x) + sin(x)cos(x) = 0[/tex]

[tex]cos(x)( - cos(x) + sin(x)) = 0[/tex]

[tex]cos(x) = 0 \\ x = \frac{\pi}{2} + k\pi \\ \\ sin(x) = cos(x) \\ x = \frac{\pi}{4} + k\pi[/tex]

Answer:

a) x = 1.5 and x = -0.3

b) x = -8 and x = 5

Step-by-step explanation:

a)

The given equation follows the general structure: ax² + bx + c = 0.

Therefore, if a = 5, b = -6, and c = -2, you can substitute the values into the quadratic formula and solve for "x".

b)

Another way of solving polynomials is through factorization. After rearranging the equation to fit the general structure of a quadratic (as seen above), you can factor by asking yourself the question, which 2 numbers multiply to "c" (-40) and add to "b" (3)? The answers will make up your factors.

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\huge\textbf{Equation \#1. }[/tex]

[tex]\mathsf{5x^2 - 6x - 2 = 0}[/tex]

[tex]\huge\textbf{Use the quadratic formula to solve:}[/tex]

[tex]\mathsf{x = \dfrac{-(-6)\pm \sqrt{(-6)^2 - 4(5)(-2)}}{2(5)}}[/tex]

[tex]\huge\textbf{Simplify it: }[/tex]

[tex]\mathsf{x = \dfrac{6 \pm \sqrt{76}}{10}}[/tex]

[tex]\huge\textbf{Simplify that as well:}[/tex]

[tex]\mathsf{x = \dfrac{3}{5} + \dfrac{1}{5}\sqrt{19}\ or\ x = \dfrac{3}{5} + (-\dfrac{1}{5})\sqrt{19}}[/tex]

[tex]\huge\textbf{Therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\mathsf{x \approx 1.5 \ or\ x\approx -0.3{}\ }}\huge\checkmark[/tex]

[tex]\huge\textbf{Equation \#2.}[/tex]

[tex]\mathsf{x^2 + 3x = 40}[/tex]

[tex]\huge\textbf{Subtract 40 to both sides:}[/tex]

[tex]\mathsf{x^2 + 3x - 40 = 40 - 40}[/tex]

[tex]\huge\textbf{Simplify it:}[/tex]

[tex]\mathsf{x^2+ 3x - 40 = 0}[/tex]

[tex]\huge\textbf{Factor the left side of the equation:}[/tex]

[tex]\mathsf{(x - 5)\times (x + 8) = 0}[/tex]

[tex]\mathsf{(x - 5)(x + 8) = 0}[/tex]

[tex]\huge\textbf{Set the factors to equal to 0:}[/tex]

[tex]\mathsf{x - 5 = 0 \ or\ even\ x + 8 = 0}[/tex]

[tex]\huge\textbf{Simplify it:}[/tex]

[tex]\mathsf{x = 5\ or\ x = -8}[/tex]

[tex]\huge\textbf{Therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\mathsf{x = 5\ or \ x = -8}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

Step-by-step explanation:

[tex]52 = - 5x - 3i...given \: expression \\ - 5x - 3i = 52...switch \: sides \\ - 5x - 3i + 3i = 52 + 3i...add \: 3i \: to \: both \: sides \\ - 5x = 52 + 3i...simplify \\ \frac{ - 5x}{ - 5} = \frac{52}{ - 5} + \frac{3i}{ - 5} ...divide \: both \: sides \: by \: - 5 \\ x = \frac{ - 52}{5} - i \frac{3}{5} ...simplified[/tex]

Answer:

The statement is false.

Step-by-step explanation:

Given,

[tex] \sqrt{4 \times - 3} = 5[/tex]

To Prove

Soln:

=[tex]2i \sqrt{3} [/tex]

=>[tex]2i \sqrt{3} ≠5[/tex]

Hence, 2i√3 is not equal (≠) to 5.

The system of equations to the inverse of its coefficient matrix, A⁻¹, and the matrix of its solution, X is shown in the figure.

Given that the system of equations are shown in given figure.

The first system of equations are

[tex]\begin{aligned}4x+2y-z&=150\\x+y-z&=-100\\-3x-y+z&=600\\\end[/tex]

By writing in matrix AX=b, we get

Coefficient matrix [tex]A=\left[\begin{array}{lll}4&2&-1\\1&1&-1\\-3&-1&1\end{array}\right][/tex] and [tex]B=\left[\begin{array}{l}150&-100&600\end{array}\right][/tex]

Firstly, we will find the A⁻¹ by finding the determinant and adjoint of A and divide the adjoint with determinant, we get

[tex]\begin{aligned}|A|&=\left|\begin{array}{lll}4&2&-1\\1&1&-1\\-3&-1&1\end{array}\right|\\ &=4(1-1)-2(1-3)-1(-1+3)\\&=4(0)-2(-2)-1(2)\\ &=2\neq 0\end[/tex]

[tex]\begin{aligned}Adj A&=\left[\begin{array}{lll}0&2&2\\-1&1&-2\\-2&3&2\end{array}\right]^T\\&=\left[\begin{array}{lll}0&-1&-2\\2&1&3\\2&-2&2\end{array}\right]\end[/tex]

[tex]\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0&-0.5&-0.5\\1&0.5&1.5\\1&-1&1\end{array}\right]\end[/tex]

For a solution Consider [A B] and apply row operations, we get

[tex]\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{lll1}4&2&-1&150\\1&1&-1&-100\\-3&-1&1&600\end{array}\right]\\ R_{2}&\rightarrow 4R_{2}-R_{1},R_{3}\rightarrow 4R_{3}+3R_{1}\\ &\sim \left[\begin{array}{lll1}4&2&-1&150\\0&2&-3&-550\\0&2&1&2850\end{array}\right]\\ R_{3}&\rightarrow R_{3}-R_{2}\\ &\sim \left[\begin{array}{llll}4&2&-1&150\\0&2&-3&-550\\0&0&4&3400\end{array}\right]\end[/tex]

Thus, [tex]x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}-250\\1000\\850\end{array}\right][/tex]

The second system of equations are

[tex]\begin{aligned}x+y-z&=220\\5x-5y-z&=-640\\-x+y+z&=200\\\end[/tex]

Similarly, we will find for second system of equations

[tex]\begin{aligned}|A|&=\left|\begin{array}{lll}1&1&-1\\5&-5&-1\\-1&1&1\end{array}\right|\\ &=1(-5+1)-1(5-1)-1(5-5)\\&=1(-4)-1(4)-1(0)\\ &=-8\neq 0\end[/tex]

[tex]\begin{aligned}Adj A&=\left[\begin{array}{lll}-4&-4&0\\-2&0&-2\\-6&-4&-10\end{array}\right]^T\\&=\left[\begin{array}{lll}-4&-2&-6\\-4&0&-4\\0&-2&-10\end{array}\right]\end[/tex]

[tex]\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0.5&0.25&0.75\\0.5&0&0.5\\0&0.25&1.25\end{array}\right]\end[/tex]

[tex]\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{llll}1&1&-1&220\\5&-5&-1&-640\\-1&1&1&200\end{array}\right]\\ R_{2}&\rightarrow R_{2}-5R_{1},R_{3}\rightarrow R_{3}+R_{1}\\ &\sim \left[\begin{array}{llll}1&1&-1&220\\0&-10&4&-1740\\0&2&0&420\end{array}\right]\\ R_{3}&\rightarrow 5R_{3}+R_{2}\\ &\sim \left[\begin{array}{llll}1&1&-1&220\\0&-10&4&-1740\\0&0&4&360\end{array}\right]\end[/tex]

Thus, [tex]x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}100\\210\\90\end{array}\right][/tex]

The third system of equations are

[tex]\begin{aligned}2x+2y-z&=290\\x+y-3z&=500\\x-y+2z&=600\\\end[/tex]

Similarly, we will find for third system of equations

[tex]\begin{aligned}|A|&=\left|\begin{array}{lll}2&2&-1\\1&1&-3\\1&-1&2\end{array}\right|\\ &=2(2-3)-2(2+3)-1(-1-1)\\&=2(-1)-2(5)-1(-2)\\ &=-10\neq 0\end[/tex]

[tex]\begin{aligned}Adj A&=\left[\begin{array}{lll}-1&-5&-2\\-3&5&4\\-5&5&0\end{array}\right]^T\\&=\left[\begin{array}{lll}-1&-3&-5\\-5&5&5\\-2&4&0\end{array}\right]\end[/tex]

[tex]\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0.1&0.3&0.5\\0.5&-0.5&-0.5\\0.2&-0.4&0\end{array}\right]\end[/tex]

get

[tex]\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{llll}2&2&-1&290\\1&1&-3&500\\1&-1&2&600\end{array}\right]\\ R_{2}&\rightarrow 2R_{2}-R_{1},R_{3}\rightarrow 2R_{3}-R_{1}\\ &\sim \left[\begin{array}{llll}2&2&-1&290\\&0&-5&710\\0&-4&5&910\end{array}\right]\end[/tex]

Thus, [tex]x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}479\\-405\\-142\end{array}\right][/tex]

Hence, each system of equations to the inverse of its coefficient matrix, A⁻¹, and the matrix of its solution, X.

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Answer:

- 226

Step-by-step explanation:

ARITHMETIC SEQUENCE.

Number of term of an Arithmetic progressions has the formular.

Tn = a + ( n - 1 ) d

From the question,

First term ( a ) = 70

common difference = T2 - T1 = 63 - 70 = -7

For the 49th term

T49 = a + 48d

= 70 + 48 ( -7 )

= 70 - 336 = - 226

Answer:

  C.  f(x) = 2x⁴ +6x³ +22x² +54x +36

Step-by-step explanation:

You can use Descartes' rule of signs and the y-intercept to help you select the correct answer.

The given point (0, 36) is the y-intercept of the function. This tells you 36 is the constant in the polynomial, eliminating choices A and B.

Descartes' rule of signs tells you the number of positive real roots will be less than or equal to the number of sign changes in the coefficients when the function is written in standard form. The number of negative real roots will be the number of sign changes after the signs of odd-degree terms are reversed.

The given real roots are both negative. There are zero positive real roots, so all of the signs of the coefficients in the function must be the same (no changes). This eliminates choice D, and tells you C is the correct answer.

  f(x) = 2x⁴ +6x³ +22x² +54x +36

Answer:

In 3h (12am)

Step-by-step explanation:

First car A Will by 10 am be 40 miles from staring point, then car B Will start going 60mph and by 11am car A Will be 80 miles from start, and car B Will be 60 miles from start in 12 am car A Will be at 120 miles and car B Will be also 120 miles

And answer is in 3h or in 12am

The domain of the function g(x) = –⌊x⌋ + 3 is (a) {x| x is a real number}

The function is given as

g(x) = –⌊x⌋ + 3

The above is a step function, and the domain is the set of input values it can accept

Step functions of the given form can accept any real value of x

Hence, the domain of the function g(x) = –⌊x⌋ + 3 is (a) {x| x is a real number}

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Complete question

The graph of the step function g(x) = –⌊x⌋ + 3 is shown. What is the domain of g(x)? {x| x is a real number} {x| x is an integer} {x| –2 ≤ x < 5} {x| –1 ≤ x ≤ 5}

Answer:

A

Step-by-step explanation:

Answer:

172.5cm^3

Step-by-step explanation:

* = multiply or times

volume = length*width*height

Plug in the numbers: 11.5*2.5*6 = 172.5cm^3

Following are the dependent variables:

1. The amount of water that each orchard receives.

2. The species of trees in the orchard.

Reason:

The exercise scientist is looking for the effects of a chemical between an apple crop to which it is administered and another to which it is not, 4 options are presented, of which it is essential to count as a variable the amount of water each Orchard and tree species in the orchard, since they can generate alterations in the results, the other two variables of the exercise such as number of apples and size of the orchards are not significant and their variations do not affect the scientist's objective.

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The equation which represents the equation for the temperature, D , in terms of t is D(t)=5°cos{(π/3)t}+67°.

Given that the high temperature is 72 degrees and low temperature is 62 degrees at 3 A.M.

We know that temperature is the intensity of the heat present around us.

We know that,

Maximum temperature=72 degrees,

Minimum temperature=62 degrees, which occurs at t=3 hours

Now we can write the equation as:

D(t)=A cos(ct)+B

Where A, c, B are constants.

We have a minimum at t=3 a minimum means cos(ct)=-1

then we have that D(3)=A cos(c*3)+B

=A*(-1)+b

=35°

Here we solve that ,

Cos(c*3)=-1

this means that

c*3=-1

c*3=π

c=π/3

We also know that the maximum temperature is 72°, the maximum temperature is when cos(c*t)=1

D(t)=0=A(t)+B=72

With this we can find that values of A and b

-A+B=62

A+B=72

B=67

A=5

Equation will be D(t)=5 cos{(π/3)t}+67°.

Hence the equation for the temperature is D(t)=5 cos{(π/3)t}+67°..

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Answer:

2700

Step-by-step explanation:

*6*10*9/4!*4*4*6*5/4!

The type of bacterial pneumonia is most often seen in children and young adults is; Pneumococcal infection.

The correct type of Bacterial pneumonia that is most often seen in children and young adult is called Pneumococcal infection. This is because it is a name for any infection caused by bacteria called Streptococcus pneumoniae, or pneumococcus.

Thus, we can conclude that the type of Bacterial pneumonia here is called Pneumococcal infection.

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Answer:

The gradient is -[tex]\frac{2}{3}[/tex] and the intercept is 2.

Step-by-step explanation:

First, transform the given equation into the slope-intercept form, y = mx + b. The variable m represents slope (gradient) and the variable b represents the intercept.

2x + 3y = 6

3y = -2x + 6

y = -[tex]\frac{2}{3}[/tex]x + 2

The gradient is -[tex]\frac{2}{3}[/tex] and the intercept is 2.

Answer:

[tex]\sf gradient =\dfrac{-2}{3}\\\\ y -intercept = 2[/tex]

Step-by-step explanation:

      [tex]\sf \boxed{\bf y = mx +b}[/tex]

Here m is the slope or gradient and b is the y-intercept.

Write the given equation in slope-intercept form.

  2x + 3y = 6

          3y = -2x + 6

            [tex]\sf y =\dfrac{-2}{3}x +\dfrac{6}{3}\\\\ y = \dfrac{-2}{3}x + 2[/tex]

[tex]\sf gradient =\dfrac{-2}{3}\\\\ y -intercept = 2[/tex]

Formula: U*V=R*T

3*1=4*x

3=4x

x=3/4

Hope it helps!

Answer:

[tex]\sf PR =\dfrac{3}{4}[/tex]

Step-by-step explanation:

It two chords or secants intersect inside the circle, then the product of the length of the segments of one chord is equal to the product of the lengths of the segments of the other chords.

 TP * PR = UP * PV

    4 * PR = 3 * 1

          [tex]\sf PR = \dfrac{3}{4}[/tex]

Step-by-step explanation:

we have

2y = 4x - 9

and we want it to look like

...x + ...y = -9

simple.

the y term is already on the left side. we need to move the x term to the same side.

what do we do ? we subtract the term we want to get rid of on one side from both sides (we always have to do changes in both sides of the equation, or we change the whole meaning of the equation).

2y = 4x - 9 | -4x on both sides

-4x + 2y = -9

and we are finished. that's it.

Answer:

-4x + 2y = -9

Step-by-step explanation:

We are given the equation 2y=4x-9, and we want to convert it into standard form.

Standard form is written as ax+by=c, where a, b, and c are free integer coefficients, however a and b cannot be 0.

Notice how in standard form, x and y are on the same side. Currently, x and y are on different sides.

Therefore, we first need to get x and y on the same side.

We can do this by subtracting 4x from both sides.

2y = 4x - 9

-4x   -4x

_____________

-4x + 2y = -9

As indicated by the -9 on the left side, we have solved the question, and are now done.

Hence, the answer is -4x + 2y = -9.

Part 1

Angles in a triangle add to 180 degrees, so

[tex]m\angle E=180^{\circ}-82^{\circ}-43^{\circ}=55^{\circ}[/tex]

Part 2

By the Law of Sines,

[tex]\frac{EF}{\sin 43^{\circ}}=\frac{15}{\sin 55^{\circ}}\\\\EF=\frac{15 \sin 43^{\circ}}{\sin 55^{\circ}}\\\\EF \approx 12.49[/tex]

Answer:

What is the measure of angle E?

m∠E =

✔ 55

°

What is the length of EF rounded to the nearest hundredth?

EF ≈

✔ 12.49

Step-by-step explanation:

Answer:

(11a^5 - 3b^4) (3b^4 + 11a^5) = 121a^10 - 9b^8

Step-by-step explanation:

I got the 11 part by doing √121 = 11

I got the a^5 by knowing that a needs to have the same exponent both times and 5+5=10. Thats how I got the a^5 part.

Answer: (11a^5 - 3b^4) (3b^4 + 11a^5) = 121a^10 - 9b^8

The equation which represents the number sentence given in the task content is; n/3 + 9 = 21.

According to the task content, it follows that the sentence given is; Nine more than the quotient of a number and 3 is 21.

Since, the quotient of a number and 3 can be written as; n/3.

Consequently, the correct equation is; n/3 +9 = 21.

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Answer:

3x + 2 >= 0

Step-by-step explanation:

Since this is a 4th root, not a cubic root, the radical can only contain 0 or positive numbers. Therefore there is 3x + 2 >= 0.

The probability that the child has measles is gotten as; 0.35

F is the event of a child being sick with flu.

M is the event of a child being sick with measles.

A is the event that the doctor finds a rash.

B1 is the event that the child has measles

S is the sick children.

P(R|M) = 0.93.

P(R|F) = 0.09

P(S|F) = 0.95

P(S|M) = 0.05

Thus, the probability that the child has measles is;

P(M|R) = [(0.05 * 0.93)/[(0.05 * 0.93) + (0.95 * 0.09)]

P(M|R) = 0.35

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Using the Empirical Rule, it is found that the interval in which approximately 978 students scored was:

A. 72 < x < 88.5.

It states that, for a normally distributed random variable:

The percentage that 978 is of 1200 is:

978/1200 x 100% = 81.5%.

Considering the symmetry of the normal distribution, two outcomes are possible involving 81.5% of the measures:

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