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2 years ago

15

What is the length of EF in the right triangle below?

2 answers:

Lyrx [107]2 years ago

8 0

Answer:

\sqrt(217)

Step-by-step explanation:

(19)^2=(12)^2

361 - 144 + 217

\sqrt(217)

andriy [413]2 years ago

8 0

Answer:

$\sqrt{217}units$

Step-by-step explanation:

Given : A right angled triangle EFD

Hypotenuse = ED=19 units

Base = DF =12 units

Perpendicular = EF

Solution:

To find length of EF we will use Pythagorean Theorem:

$(Hypotenuse)^{2}=(Perpendicular)^{2}+(Base)^{2}$

$ED^{2}=EF^{2}+DF^{2}$

$19^{2}=EF^{2}+12^{2}$

$361=EF^{2}+144$

$361-144=EF^{2}$

$217=EF^{2}$

$\sqrt{217} =EF$

Thus the length of EF is $\sqrt{217}units$

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3 lines are shown. A line with points M, H, K intersects with a line with points J, H, L at point H. Another line extends from p

otez555 [7]

Answer:

Option B.

Step-by-step explanation:

Given information: ∠MHL=(3x+20), ∠KHN=(x+25), and ∠JHN=(x+20).

We need to find the measure of ∠JHN.

$\angle MHL=\angle JHK$ (Vertical opposite angles)

$\angle MHL=\angle JHN+\angle KHN$

Substitute the given values.

$3x+20=(x+20)+(x+25)$

$3x+20=2x+45$

$3x-2x=45-20$

$x=25$

The value of x is 25. So, the measure of ∠JHN is

$\angle JHN=x+20=25+20=45$

The measure of ∠JHN is 45°.

Therefore, the correct option is B.

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2 years ago

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If f(x) = (xm + 9)2, which statement about f(x) is true?

Marina CMI [18]

Answer:

The correct statement for the given function is:

f(x) is an even function for all even values of m.

Step-by-step explanation:

The correct statement for the given function is:

f(x) is an even function for all even values of m.

The given function(x^m+9)^2 is a quadratic equation. The term Quadratic means the variable gets squared. if m is an even number, that would result also to an even function....

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2 years ago

Complete the steps to find all zeroes of the function f(x) = x4 − 4x3 − 4x2 + 36x − 45. This function has roots.

xxTIMURxx [149]

Steps?

A graph shows zeros to be ±3. Factoring those out leaves the quadratic
(x-2)² +1
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2 years ago

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Mr. Lobo is building a fence rectangular around his rectangular yard. It is 16 feet long and 14 feet wide. How much feet of feet

lawyer [7]

Answer:

30

Step-by-step explanation:

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4 0

2 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago