Help me pls asappp with Pythagorean theorem

Option C

Corresponding angles along parrellel lines are conguerent

Answered by Gauthmath pls mark brainliest and comment thanks and click thanks

Answer:

a.) 19.2cm

b.) 0.15375cm

Step-by-step explanation:

Cylinders are similar, so:

h1 / r1 = h2 / r2

8cm / 5cm = h2 / 12cm

h2 = (8cm × 12cm) / 5cm

h2 = 19.2cm

Same for b

32000cm2 / 246cm = 20cm2 / length

length = ((20 × 246) / 32000) cm

length = 0.15375cm

Step-by-step explanation:

A ball is thrown straight up from a rooftop 320 feet high. The formula below describes the ball's height above the ground, h, in feet, t seconds after it was thrown. The ball misses the rooftop on its way down and eventually strikes the ground. How long will it take for the ball to hit the ground? Use this information to provide tick marks with appropriate numbers along the horizontal axis in the figure shown.

h=-16t^2+16t+320

Answer:

yes the x increases by 6 and the y decreases by 3.

y = -1/2x - 7/2

Step-by-step explanation:

find the slope :

(1,-4), (7, -7)

y2- y1 / x2 - x1

substitute those numbers and you get -1/2.

point slope form :

y - y1 = m(x- x1)

y - (-4) = -1/2 ( x - (1))

y+4 = -1/2(x-1)

slope intercept form :

y = -1/2x - 7/2

does this help ?

Answer:

8-x=2

-x=2-8

-x=-6

x=6

if 8 is subtract from 6answer is 2

Answer:

-10k×2+7

= -20k+7

Step-by-step explanation:

is the answer

Answer:

hey the answer is cylinder= 211.5л!

Answer:

um 1/4

Step-by-step explanation:

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Whenever she began to drive farther east, at 70 km per hour, Prachi was 5 km east of her home.

Let f(n) at the beginning of her nth hour drive be Prachi's distance to her home.

f is a sequence of arithmetic.

Write a series explicit form.

[tex]\to f(n) = 70n + 5[/tex]

Answer:

Arithmetic Sequence, f(n)=5+70(n-1)

Step-by-step explanation:

checked on khan

====================================================

Work Shown:

Plug in x = 0

[tex]f(x) = \frac{2x-1}{3x+5}\\\\f(0) = \frac{2*0-1}{3*0+5}\\\\f(0) = \frac{0-1}{0+5}\\\\f(0) = -\frac{1}{5}\\\\[/tex]

Repeat for x = 8

[tex]f(x) = \frac{2x-1}{3x+5}\\\\f(8) = \frac{2*8-1}{3*8+5}\\\\f(8) = \frac{16-1}{24+5}\\\\f(8) = \frac{15}{29}\\\\[/tex]

Now use the average rate of change formula

[tex]m = \frac{f(b)-f(a)}{b-a}\\\\m = \frac{f(8)-f(0)}{8-0}\\\\m = \frac{15/29 - (-1/5)}{8}\\\\m = \frac{15/29 + 1/5}{8}\\\\m = \frac{(15/29)*(5/5) + (1/5)*(29/29)}{8}\\\\m = \frac{75/145 + 29/145}{8}\\\\[/tex]

[tex]m=\frac{104/145}{8}\\\\m = \frac{104}{145} \div \frac{8}{1}\\\\m = \frac{104}{145} \times \frac{1}{8}\\\\m = \frac{104*1}{145*8}\\\\m = \frac{104}{1160}\\\\m = \frac{13}{145}\\\\[/tex]

please mark this answer as brainlist

Answer:

Shannon

Step-by-step explanation:

Cuando se habla de lo que uno tiene, podemos usar números positivos.

Por ejemplo:

Pedro tiene 10 manzanas.

Para el caso de deudas, utilizamos números negativos, por ejemplo:

Pedro tiene -10 manzanas

Lo cual significa que Pedro debe 10 manzanas a alguien.

Entonces si le diéramos a Pedro 12 manzanas, el ahora tendría:

-10 + 12 = 2

Pedro tiene 2 manzanas, porque tuvo que entregar 10 de las 12 que le dimos para pagar su deuda.

Ahora vamos a resolver el problema:

La cuenta de Sally tiene un saldo de:

S = -$200.90

El signo negativo quiere decir que Sally tiene una deuda de $200.90

La cuenta de su amiga Shannon tiene un saldo de:

S' = -$240.55

De vuelta, el signo negativo quiere decir que Shannon tiene una deuda de $240.55

Con esto ya podemos concluir que la deuda de Shannon es mayor, por lo tanto Shannon es la que tiene más deuda.

Answer:

There are no real solutions.

Step-by-step explanation:

There are 3 options.

2 real solutions: This happens if in the graph, each arm intersects the x-axis, this means that there are two different values of x such that the equation:

a*x^2 + b*x + c

is equal to zero.

Another way to see this, is if the determinant:

b^2 - 4*a*c

is larger than zero.

1 real solution: This happens when the vertex of the graph intersects the x-axis. This means that there is a single value of x such that:

a*x^2 + b*x + c

is equal to zero.

Another way to see this is if the determinant:

b^2 - 4*a*c

is larger equal zero.

No real solution: if in the graph we can not see any intersection of the x-axis, then we do not have real solutions (only complex ones).

Another way to see this is if the determinant:

b^2 - 4*a*c

is smaller than zero.

Now that we know this, let's look at the graph.

We can see that the vertex is below the x-axis, and the arms of the graph go downwards. So the arms will never intersect the x-axis (and neither the vertex).

So the graph does not intersect the x-axis at any point, which means that there are no real solutions for the quadratic equation.

The correct answer would be "none"

Answer:

side A should be about 44cm

Answer:

DE = 21.4

Step-by-step explanation:

The parallel lines divide the transversals proportionally, that is

[tex]\frac{DE}{EB}[/tex] = [tex]\frac{DF}{FC}[/tex] , substitute values

[tex]\frac{DE}{10.7}[/tex] = [tex]\frac{32}{16}[/tex] = 2 ( multiply both sides by 10.7 )

DE = 21.4

Answer:

8x

Step-by-step explanation:

=3x+5x

=8x

Answer:

Step-by-step explanation:

The series is missing from the question. I will answer this question with a general explanation by using the following similar series:

[tex]\sum\limits^6_{n=0} 7^n[/tex]

Required

The series

To do this, we simply replace n with the values

[tex]\sum\limits^6_{n=0}[/tex] means n starts from 0 and ends at 6

[tex]\sum[/tex] means the series is a summation series

So, we have:

[tex]\sum\limits^6_{n=0} 7^n = 7^0 + 7^1 + 7^2 + ...... + 7^6[/tex]

[tex]\sum\limits^6_{n=0} 7^n = 1 + 7 + 7^2 + ...... + 7^6[/tex]

Answer with explanation:

Yes, each point on a circle is a fixed distance from the center of the circle. This is called the radius of the circle. By definition, all radii of a circle are equal.

Similar polygons have corresponding sides in similar proportion. Regardless of how large a circle is, each point on the circle will still be a fixed distance away from center of the circle. Therefore, the radii are in a constant proportional and all circles are similar.

Answer:

[tex]{ \tt{(2x + y)( {n}^{2} - 3xy + {y}^{2} )}} \\ = { \tt{(2x {n}^{2} - 3 {x}^{2}y + 2x {y}^{2} + {n}^{2}y - 3x {y}^{2} + {y}^{3} )}} \\ = { \tt{ {y}^{3} - xy(y + 3x) + {n}^{2} y }}[/tex]

Hello!

(2x+y) (n2-3xy+y2)

2x* n²= 2xn²

2x* -3xy = -6x²y

2x* y² = 2xy²

y*n² = yn²

y*-3xy = -3xy²

y* y² = y³

=>

2xn²- 6x²y + 2xy² +yn²- 3xy² +y³

Answer:

The function that represents g(x) is the third choice: g(x) = (x − 4)^2 + 9

Step-by-step explanation:

The original function has been shifted 9 units up (a vertical transformation). To show a vertical transformation, all we have to do is either add or subtract at the end of the function.

To show a shift upwards, we add the value of change.

To show a shift downwards, we subtract the value of change.

In this case, the original function f(x) = [tex]x^{2}[/tex] was translated 9 units up. Since we shifted up, we simply add 9 to the end of the function: g(x) = [tex]x^{2}[/tex] + 9

The original function has also been shifted 4 units to the right. This is a horizontal transformation. To show a horizontal transformation, we need to either add or subtract within the function (within the parenthesis).

To show a shift to the left, we add the value of change.

To show a shift to the right, we subtract the value of change.

*Notice: Moving left does NOT mean to subtract while moving right does NOT mean to add. The rules above are counterintuitive so pay attention when doing horizontal transformations.

In this case, the original function f(x) = [tex]x^{2}[/tex] was translated 4 units to the right. Since we shifted right, we must subtract 4 units within the function/parenthesis: g(x) = [tex](x-4)^{2}[/tex]

When we combine both vertical and horizontal changes, the only equation that follows these rules is the third choice: g(x) = (x − 4)^2 + 9

Answer: C

Step-by-step explanation:

Answer:

See Below (Boxed Solutions).

Step-by-step explanation:

We are given the two complex numbers:

[tex]\displaystyle z = \sqrt{3} - i\text{ and } w = 6\left(\cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12}\right)[/tex]

First, convert z to polar form. Recall that polar form of a complex number is:

[tex]z=r\left(\cos \theta + i\sin\theta\right)[/tex]

We will first find its modulus r, which is given by:

[tex]\displaystyle r = |z| = \sqrt{a^2+b^2}[/tex]

In this case, a = √3 and b = -1. Thus, the modulus is:

[tex]r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2[/tex]

Next, find the argument θ in [0, 2π). Recall that:

[tex]\displaystyle \tan \theta = \frac{b}{a}[/tex]

Therefore:

[tex]\displaystyle \theta = \arctan\frac{(-1)}{\sqrt{3}}[/tex]

Evaluate:

[tex]\displaystyle \theta = -\frac{\pi}{6}[/tex]

Since z must be in QIV, using reference angles, the argument will be:

[tex]\displaystyle \theta = \frac{11\pi}{6}[/tex]

Therefore, z in polar form is:

[tex]\displaystyle z=2\left(\cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6}\right)[/tex]

Part A)

Recall that when multiplying two complex numbers z and w:

[tex]zw=r_1\cdot r_2 \left(\cos (\theta _1 + \theta _2) + i\sin(\theta_1 + \theta_2)\right)[/tex]

Therefore:

[tex]\displaystyle zw = (2)(6)\left(\cos\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right) + i\sin\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right)\right)[/tex]

Simplify. Hence, our polar form is:

[tex]\displaystyle\boxed{zw = 12\left(\cos\frac{9\pi}{4} + i\sin \frac{9\pi}{4}\right)}[/tex]

To find the complex form, evaluate:

[tex]\displaystyle zw = 12\cos \frac{9\pi}{4} + i\left(12\sin \frac{9\pi}{4}\right) =\boxed{ 6\sqrt{2} + 6i\sqrt{2}}[/tex]

Part B)

Recall that when raising a complex number to an exponent n:

[tex]\displaystyle z^n = r^n\left(\cos (n\cdot \theta) + i\sin (n\cdot \theta)\right)[/tex]

Therefore:

[tex]\displaystyle z^{10} = r^{10} \left(\cos (10\theta) + i\sin (10\theta)\right)[/tex]

Substitute:

[tex]\displaystyle z^{10} = (2)^{10} \left(\cos \left(10\left(\frac{11\pi}{6}\right)\right) + i\sin \left(10\left(\frac{11\pi}{6}\right)\right)\right)[/tex]

Simplify:

Simplify using coterminal angles. Thus, the polar form is:

[tex]\displaystyle \boxed{z^{10} = 1024\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)}[/tex]

And the complex form is:

[tex]\displaystyle z^{10} = 1024\cos \frac{\pi}{3} + i\left(1024\sin \frac{\pi}{3}\right) = \boxed{512+512i\sqrt{3}}[/tex]

Part C)

Recall that:

[tex]\displaystyle \frac{z}{w} = \frac{r_1}{r_2} \left(\cos (\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right)[/tex]

Therefore:

[tex]\displaystyle \frac{z}{w} = \frac{(2)}{(6)}\left(\cos \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right) + i \sin \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right)\right)[/tex]

Simplify. Hence, our polar form is:

[tex]\displaystyle\boxed{ \frac{z}{w} = \frac{1}{3} \left(\cos \frac{17\pi}{12} + i \sin \frac{17\pi}{12}\right)}[/tex]

And the complex form is:

[tex]\displaystyle \begin{aligned} \frac{z}{w} &= \frac{1}{3} \cos\frac{5\pi}{12} + i \left(\frac{1}{3} \sin \frac{5\pi}{12}\right)\right)\\ \\ &=\frac{1}{3}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right) + i\left(\frac{1}{3}\left(- \frac{\sqrt{6} + \sqrt{2}}{4}\right)\right) \\ \\ &= \boxed{\frac{\sqrt{2} - \sqrt{6}}{12} -\frac{\sqrt{6}+\sqrt{2}}{12}i}\end{aligned}[/tex]

Part D)

Let a be a cube root of z. Then by definition:

[tex]\displaystyle a^3 = z = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]

From the property in Part B, we know that:

[tex]\displaystyle a^3 = r^3\left(\cos (3\theta) + i\sin(3\theta)\right)[/tex]

Therefore:

[tex]\displaystyle r^3\left(\cos (3\theta) + i\sin (3\theta)\right) = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]

If two complex numbers are equal, their modulus and arguments must be equivalent. Thus:

[tex]\displaystyle r^3 = 2\text{ and } 3\theta = \frac{11\pi}{6}[/tex]

The first equation can be easily solved:

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

For the second equation, 3θ must equal 11π/6 and any other rotation. In other words:

[tex]\displaystyle 3\theta = \frac{11\pi}{6} + 2\pi n\text{ where } n\in \mathbb{Z}[/tex]

Solve for the argument:

[tex]\displaystyle \theta = \frac{11\pi}{18} + \frac{2n\pi}{3} \text{ where } n \in \mathbb{Z}[/tex]

There are three distinct solutions within [0, 2π):

[tex]\displaystyle \theta = \frac{11\pi}{18} , \frac{23\pi}{18}\text{ and } \frac{35\pi}{18}[/tex]

Hence, the three roots are:

[tex]\displaystyle a_1 = \sqrt[3]{2} \left(\cos\frac{11\pi}{18}+ \sin \frac{11\pi}{18}\right) \\ \\ \\ a_2 = \sqrt[3]{2} \left(\cos \frac{23\pi}{18} + i\sin\frac{23\pi}{18}\right) \\ \\ \\ a_3 = \sqrt[3]{2} \left(\cos \frac{35\pi}{18} + i\sin \frac{35\pi}{18}\right)[/tex]

Or, approximately:

[tex]\displaystyle\boxed{ a _ 1\approx -0.4309 + 1.1839i,} \\ \\ \boxed{a_2 \approx -0.8099-0.9652i,} \\ \\ \boxed{a_3\approx 1.2408-0.2188i}[/tex]

Answer: 12

Step-by-step explanation:

16 X 20c = 3.20

12 x 5c = 0.60

total is 3.80

Amy has 12 five c coins.

Answer:

[tex]10^{-3}[/tex]

Step-by-step explanation:

Answer:

https://tex.z-dn.net/?f=10%5E%7B-3%7D

Step-by-step explanation:

Answer:

Answer:

hey mate !!!

The pattern followed is

4x2-2= 8-2=6

6 x 2-3= 9

9 x 2- 4= 14

14 x 2-5 = 23

23 x 2-6= 40

40 x 2-7= 73

So, the next number will be 73.

Answer:

c

Step-by-step explanation: