CORRECT answer gets brainliest.

Jayden and Sheridan both tried to find the missing side of the right triangle. A right triangle is shown. One leg is labeled as

stellarik [79]

Answer:

Sheridan's Work is correct

Step-by-step explanation:

we know that

The lengths side of a right triangle must satisfy the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$

where

a and b are the legs

c is the hypotenuse (the greater side)

In this problem

Let

$a=7\ cm\\c=13\ cm$

substitute

$13^{2}=7^{2}+b^{2}$

Solve for b

$169=49+b^{2}$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

we have that

Jayden's Work

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\b=13\ cm$

substitute and solve for c

$7^{2}+13^{2}=c^{2}$

$49+169=c^{2}$

$218=c^{2}$

$c=\sqrt{218}\ cm$

$c=14.76\ cm$

Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle

Sheridan's Work

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\c=13\ cm$

substitute

$7^{2}+b^{2}=13^{2}$

Solve for b

$49+b^{2}=169$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

therefore

Sheridan's Work is correct

6 0

3 years ago

Read 2 more answers

Center: (10,-10)

vova2212 [387]

This is the standard form equation $\frac{(x - 10)^2}{100} + \frac{(y + 10)^2}{-100} = 1$

What is the ellipse?

The equation for an ellipse is typically written as x² a² + y² b² = 1. x² a² + y² b² = 1. An ellipse with its origin at the center is defined by this equation. The ellipse is stretched further in both the horizontal and vertical directions if a > b, a > b, and if b > a, b > a, respectively.

The standard form of the equation of an ellipse with center (h, k)and major axis parallel to the x-axis is:

$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$

where,

a > b

the length of the major axis is 2a

the coordinates of the vertices are (h±a,k)

the length of the minor axis is 2b

the coordinates of the co-vertices are (h,k±b)

the coordinates of the foci are (h±c,k),

where c² = a² − b².

so,

$\frac{(x - 10)^2}{100} + \frac{(y + 10)^2}{-100} = 1$

Hence, this is the standard form equation $\frac{(x - 10)^2}{100} + \frac{(y + 10)^2}{-100} = 1$ .

To learn more about ellipse, visit

brainly.com/question/16904744

#SPJ1

7 0

1 year ago

I ONLY need help with the tables ONLY! I’LL GIVE BRAINLIEST FOR THE RIGHT ANSWER?! I please include what you did for both tables

neonofarm [45]

Answer:

The first table.

Step-by-step explanation:

Every single time, the x is being multiplied by 3 to get the y, and that means its proportional. Also, the second one is for sure NOT proportional

7 0

3 years ago

A circle is circumscribed around a square and another circle is inscribed in the square. If the area of the square is 9 in2, wha

lesantik [10]

Answer:

√2:1

Step-by-step explanation:

First we need to know that the length of the side of the square is equal to the diameter of the inscribed circle i.e

L = di

Given the area of the square to be 9in², we can get the length of the square.

Area of a square = L²

L is the length of the square.

9 = L²

L = √9

L = 3in

Hence the length of one side of the square is 3in

This means that the diameter of the inscribed circle di is also 3in.

Circumference of a circle = π×diameter of the circle(di)

Circumference of inscribed circle = π×3

= 3π in

For the circumscribed circumscribed circle, diameter of the outer circle will be equivalent to the diagonal of the square.

To get the diagonal d0, we will apply the Pythagoras theorem.

d0² = L²+L²

d0² = 3²+3²

d0² = 9+9

d0² = 18

d0 = √18

d0 = √9×√2

d0 = 3√2 in

Hence the diameter of the circumscribed circle (d0) is 3√2 in

Circumference of the circumscribed circle = πd0

= π(3√2)

= 3√2 π in

Hence, ratio of the circumference of the circumscribed circle to the one of the inscribed will be 3√2 π/3π = √2:1

8 0

3 years ago

Find the common ratio for the following geometric sequence. 2, 1, 0.5, 0.25, . . .

Oliga [24]

The common ratio is 2:1, or dividing by 2

3 0

4 years ago

Read 2 more answers