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3 years ago

Helppp!!!! please!!!

Mathematics

2 answers:

lianna [129]3 years ago

7 0

Answer:

HEY MATE YOUR ANSWER IS

<h2>HOPE IT HELPS YOU </h2>

ch4aika [34]3 years ago

3 0

1 looks like it's going to be C but dont depend on it yet,

2 looks like A, I might be wrong sense it's been awhile.

didnt see 3 but it looks to be B.

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Pavel [41]

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4 years ago

Find the x and the area of the shape perimeter is 64.5

Umnica [9.8K]

Answer:

x=8

A = 192

Step-by-step explanation:

The perimeter is the sum of the outer lengths

17.9 + 22.6 + 3x = 64.5

Combine like terms

40.5 +3x = 64.5

Subtract 40.5 from each side

40.5 +3x -40.5 = 64.5 - 40.5

3x =24

Divide each side by 3

3x/3 = 24/3

x = 8

The area of a triangle is given by

A = 1/2 bh

The base is 3x = 3(8) =24

and the height is 2x = 2(8) =16

A = 1/2 (24) * 16

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3 years ago

Answer to the best of your abilities, pls & ty !

klemol [59]

Answer:

Step-by-step explanation:

Por ejemplo, baile

Solamente tres estudiantes siempre lo practican.

4 0

3 years ago

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In ΔJKL, the measure of ∠L=90°, KL = 22 feet, and JK = 54 feet. Find the measure of ∠J to the nearest degree.

nydimaria [60]

We have been given that in ΔJKL, the measure of ∠L=90°, KL = 22 feet, and JK = 54 feet. We are asked to find the measure of angle J to nearest degree.

First of all, we will draw a triangle as shown in the attachment.

We can see from our attachment that side KL is opposite side to angle J and side JK is hypotenuse of right triangle.

We know that sine relates opposite side of right triangle to hypotenuse.

$\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}$

$\text{sin}(\angle J)=\frac{22}{54}$

Using inverse sine or arcsin, we will get:

$\angle J=\text{sin}^{-1}(\frac{22}{54})$

$\angle J=24.042075905756^{\circ}$

Upon rounding to nearest degree, we will get:

$\angle J\approx 24^{\circ}$

Therefore, the measure of angle J is approximately 24 degrees.

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3 years ago

What is the derivative of w=(t^2+1)^100<br><br> How is it 200t(t^2+1)^99?

GuDViN [60]

(t^2+1)^100

USE CHAIN RULE

Outside first (using power rule)

100*(t^2+1)^99 * derivative of the inside

100(t^2+1)^99 * d(t^2+1)

100(t^2+1)^99 * 2t

200t(t^2+1)^99

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3 years ago

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