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tamaranim1 [39]

2 years ago

8

Help me with this, I was never taught how to do this...

1 answer:

Leya [2.2K]2 years ago

6 0

Answer:

$\sqrt{41}$ = x

Step-by-step explanation:

Basically, find the missing side lengths of each triangle so that you can solve for x:

1. right triangle w/ side lengths 48 and 52

48^2 + (missing side)^2 = 52^2
2304 + (missing side)^2 = 2704
(missing side)^2 = 400
(missing side) = 20 (use this for next triangle)

2. right triangle w/ side lengths 12 and 20 (from previous triangle)

12^2 + (missing side)^2 = 20^2
144 + (missing side)^2 = 400
(missing side)^2 = 256
(missing side) = 16 (will be used with next triangle's missing side)

3. right triangle w/ side lengths 5 and 13

5^2 + (missing side)^2 = 13^2
25 + (missing side)^2 = 169
(missing side)^2 =144
(missing side) = 12

Take right triangle w/ side lengths 12, 20, and 16 (#2) and right triangle w/ side lengths 5, 13, and 12 (#3):

the value of 12 (missing side of #3) subtracted from the #2's missing side, 16, (not its hypotenuse that equals 20 which was found from #1) will equal <u>the missing side length that will be used to find 'x'</u>
16 - 12 = 4 ---> so the side lengths are 4, 5, and x

4. Solve for the final triangle that includes x:

4^2 + 5^2 = x^2
16 + 25 = x^2
41 = x^2
$\sqrt{41}$ = x

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8 0

3 years ago

I need help please help me

Mrrafil [7]

Answer:

$y=\dfrac{1}{36}x^2$

Step-by-step explanation:

Where p is the distance from focus to vertex, the equation of the parabola with focus at (0, p) is ...

$y=\dfrac{1}{4p}x^2$

Your parabola has p=9, so the equation is ...

$y=\dfrac{1}{4\cdot 9}x^2=\bf{\dfrac{1}{36}x^2}$

3 0

3 years ago

Camille put $1,250 in a new account at her bank. - The bank pays 2.25% interest compounded annually on this account. - Camille m

Galina-37 [17]

Answer:

FV= $1,397.1

Step-by-step explanation:

Giving the following information:

Initial investment (PV)= $1,250

Interest rate (i)= 2.25% compounded annually

Number of periods (n)= 5 years

<u>To calculate the future value, we need to use the following formula:</u>

FV= PV*(1+i)^n

FV= 1,250*(1.0225^5)

FV= $1,397.1

5 0

3 years ago

What is the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y = 27 − x^2? show your

LUCKY_DIMON [66]

Answer:

Option B.

Step-by-step explanation:

The given curve is

$y=27-x^2$

We need to find the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve $y=27-x^2$ .

Let the vertex in quadrant I be (x,y), then the vertex in quadrant II is (-x,y) .

Length of the rectangle = 2x

Width of the rectangle = y

Area of a rectangle is

$Area=Length\times width$

$Area=2x\times y$

Substitute the value of y from the given equation.

$Area=2x(27-x^2)$

$A=54x-2x^3$ .... (1)

Differentiate with respect to x.

$\frac{dA}{dx}=54-6x^2$

Equate $\frac{dA}{dx}=0$ , to find the critical points.

$0=54-6x^2$

$6x^2=54$

Divide both sides by 6.

$x^2=9$

$x=\pm 3$

The value of x can not be negative because side length can not be negative.

Substitute x=3 in equation (1).

$A=54(3)-2(3)^3$

$A=162-54$

$A=108$

The area of the largest rectangle is 108 square units.

Therefore, the correct option is B.

7 0

3 years ago

cestrela7 [59]

It’s the sizes I’m pretty sure

4 0

1 year ago