Find the value of x. Round to nearest hundredth if necessary.

Answers

Answer 1

Answer: The side on the front of 30° is 1/2 of hypytenuse
Then 1/2 of 4=2 and the ather side with theorem of pytagore is squares root of(4^2-2^2)=sq .root of 12
This side(square root of 12) is 1/2 of hypotenuse then
Hypotenuse is 2*square root 12
X^2=(2*sq .root12)^2 -(square root 12)^2
X^2=48-12=36
X=6

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What is the area of a figure with vertices(1,1), (8,1) , and (5,5)?

Sqrt(15)/5 *sqrt(20)

Answers

so if you mean: $( √(15) )/(5)$ times $√(20)$ then
simplify
rememeber
$x^{ (m)/(n) }= \sqrt[n]{ x^(m) }$
and
$[x^(n)]^(m)= x^(nm)$
and
$√(x) √(y)= √(xy)$

multiply

$( √(15) )/(5)$ times $( √(20) )/(1)$ = $( √(300))/(5)$

simplify √300
√300=√3 times √100
we know that √100=10 so
√300=√3 times 10
therefor
$( √(300) )/(5)$ = $(10√(3))/(5)$ = $(5)/(5)$ times $(2√(3))/(1)$ = 1 times $(2√(3))/(1)$ =2√3

So,

$( √(15))/(5) * √(20)$

Simplify the square root of 20.
$√(20) = √(4) * √(5) = 2 * √(5) = 2 √(5)$

$( √(15))/(5) * ( 2√(5) )/(1)$

Multiply.
$( 2√(75))/(5)$

Simplify.
$( 2√(25)* √(3) )/(5)$
$( 2 * 5* √(3) )/(5)$
$(10√(3) )/(5)$
$(2√(3) )/(1)$
$2√(3)$

The length of a rectangle is 10 mm longer than its width. Its perimeter is more than 80 mm. Let w equal thewidth of the rectangle.
(a) Write an expression for the length in terms of the width.
(b) Use these expressions to write an inequality based on the given information.
(c) Solve the inequality, clearly indicating the width of the rectangle

Answers

P=L+L+W+W=2(L+W)
P>80
2(L+W)>80
divide 2
L+W>80

L is 10 more than W
L=10+W

A. L=10+W

B. L+W>80

C.
L+W>80
sub L=10+W
10+W+W>80
minus 10
2W>70
divid 2
W>35
W is mor than 35

We know that the length (L) of the rectangle in question is 7mm longer than its width (W). Let's represent that as the following:
L=7+W

A rectangle's perimeter (the total sum of its sides) will be made my 2 sides representing the length (2L) and 2 sides representing the width (2W). We also know that this rectangle's perimeter is greater than 62. Since eventually we are solving for W, let's state all expressions in terms of W:
2L=2(7+W)
2(7+W)+2W>62
14+2W+2W>62
14+4W>62
4W>62-14
4W>48
W>48/4
W>12
If the rectangle's perimeter is greater than 62, then the width will be greater than 12. Let's confirm this:
Perimeter=2L+2W
P=2(7+12)+2(12)
P=14+24+24
P=62
So we can see that if the perimeter is to surpass 62, W needs to be greater than 12 and L ( which is also 7+W) needs to be greater than 19.

PLEASE PLEASE PLEASE HURRY!Which is the best strategy to use to solve this problem?

Mr. Jackson wants to build a pool in his backyard. He plans to put a safety fence around the rectangular pool area that has an area of 400 square feet. He wants to use as little fence material as possible.

What are the dimensions Mr. Jackson should use for his rectangular pool area?

A.
Make a list.

Create a list of all possible whole-number combinations of length and width that would equal an area of 400 square feet. Then start calculating the perimeter of each rectangle. Look for a pattern to decrease the number of calculations you have to make.

B.
Write a number sentence.

Use a number sentence to calculate the area of a rectangle. Use guess and check to find two numbers that when multiplied will give a product of 400.

C.
Use objects.

Arrange 400 square tiles in different patterns until you get a rectangular shape. Count the number of tiles on the perimeter of the shape.

Answers

I believe it might be A or C. Not really sure

A cylindrical pail that has the base area of 9 pi inches squared and a height of 10 inches. One friend bought a pyramid mold with a square base with edge length of 4 inches and height of 7 inches. The other friend bought a cone with a radius of 2.5 inches and the height of six inches. What is the volume of these three objects?

Answers

Answer:

cylinder — 90π in³
pyramid — 37 1/3 in³
cone — 12.5π in³

Step-by-step explanation:

The volume of a cylinder is given by ...

V = Bh . . . . . where B is the base area and h is the height

The volume of a pyramid or cone is given by ...

V = (1/3)Bh . . . . . where B is the base area and h is the height

The area of a square of side length s is ...

A = s²

The area of a circle of radius r is ...

A = πr²

___

Using these formulas, the volumes of these objects are ...

cylinder: (9π in²)(10 in) = 90π in³

square pyramid: (1/3)(4 in)²(7 in) = 37 1/3 in³

cone: (1/3)(π(2.5 in)²)(6 in) = 12.5π in³ . . . . slightly larger than the pyramid

Answer:

12.5

Step-by-step explanation:

What is the distance between the points ( 38,30) and (14,20)?

Answers

To find out the distance of two coordinates, we can use this equation.

Distance S = [ (y2-y1)^2 + x2-x1)^2 ] ^ 0.5

x1 , y1 = 38 , 30
x2 , y2 = 14 , 20

Distance S = [ (20-30)^2 + (14-38)^2 ] ^0.5
= [ 10^2 + (-24)^2 ] ^ 0.5
= [ 100 + 576 ] ^ 0.5
= 676 ^ 0.5
= 26

(24,10)

Hope this helps!

Kevin is going to purchase sod for his backyard (see diagram below). How many square feet of sod will Kevin need?

Answers

Answer:

450 square feet

Step-by-step explanation:

Think of the shape as a complete rectangular

The area of the rectangular shape would be 25×30 = 750 square feet

Now find the area of the missing space and subtract it from 750

it would be 15×20 = 300 square feet

750 - 300 = 450 square feet

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