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2 years ago

6

Help is appreciate it is for online Class and is very confusing

1 answer:

Ksivusya [100]2 years ago

6 0

Answer:

The y=2 because the line that is going through the graph, meets the y axis at 2

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What’s the value of x?

Masteriza [31]

the answer for x is 7.27 recurring

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2 years ago

A gift in a cubic box with 12 in. edge lengths needs to be wrapped. How many square inches of wrapping paper will be needed to c

Komok [63]

864 in^2 as total surface area of the cube= 6 x side^2 = 6 x 12 x 12 in^2

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2 years ago

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A cylindrical pop can us 12.5 cenimeters tall and 5.5 centimeters wide. what is the volume of the pop can

lozanna [386]

The width is basically the diameter:
Given
Diameter = 5.5 cm
So Radius = $\frac{d}{2}$ = $\frac{5.5}{2}$ = 2.75 cm
Height of cylinder = 12.5 cm

Formula of the volume of a cylinder
V = Area of base × height of prism
= $\pi$ r² × h
=(2.75)² ( $\pi$ ) × 12.5
=7.5625 $\pi$ × 12.5
≈ 296.83 cm³

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2 years ago

List 13 meters, 15,000 meters, and 1900 centimeters in order from smallest to largest

Paha777 [63]

1 m = 100 cm

13 m is 13 m.

15 000 m is 15 000 m.

1900 cm = 1900/100 = 19m.

So we can see that 1900 cm which is 19 m is bigger than 13 m.

So from the smallest to the largest,

13m, 1900 cm, 15000 m.

Hope this explains it.

4 0

2 years ago

Given: ΔABC ≅ ΔEDF

V125BC [204]

we know that

the triangle ABC and triangle EDF are congruent triangles-----> given problem

therefore

$AC=EF\\AB=ED\\BC=DF$

we know that

the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step $1$

<u>Find the distance AB</u>

$A(0,1)\\B(0,2)$

substitute in the formula of the distance

$d=\sqrt{(2-1)^{2}+(0-0)^{2}}$

$d=\sqrt{(1)^{2}+(0)^{2}}$

$dAB=1\ unit$

Step $2$

<u>Find the distance BC</u>

$B(0,2)\\C(3,2)$

substitute in the formula of the distance

$d=\sqrt{(2-2)^{2}+(3-0)^{2}}$

$d=\sqrt{(0)^{2}+(3)^{2}}$

$dBC=3\ units$

Step $3$

<u>Find the distance AC</u>

we know that

the triangle ABC is a right triangle

so

Applying the Pythagorean Theorem

$AC^{2}=AB^{2} +BC^{2}$

substitute the values in the formula

$AC^{2}=1^{2} +3^{2}$

$AC=\sqrt{10}\ units=3.16\ units$

round to the nearest tenth

$AC=3.2\ units$

therefore

$EF=3.2\ units$

$ED=1\ unit$

$DF=3\ units$

<u>the answer is</u>

The length of the hypotenuse is equal to $3.2\ units$

4 0

2 years ago

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