I need help with this math homework!!!

What is the value of x? Enter your answer in the box. x = ?

Tema [17]

The partial circles in each corner mean that all 3 angles are identical, which means this is an equatorial triangle. In an equatorial triangle all three sides are the same.

We can set two of the equations to equal each other and solve for x:

3x -5 = 2x +20

Add 5 to each side:

3x = 2x +25

Subtract 2x from each side:

x = 25

7 0

3 years ago

A glass paperweight has a composite shape: a square pyramid fitting exacty on top of an 8 centimeter cube. The pyramid has a hei

ArbitrLikvidat [17]

Answer:

Part 1) The volume of the paperweight is $576\ cm^{3}$

Part 2) The total surface area of the paperweight is $400\ cm^{2}$

Step-by-step explanation:

Part 1) what is the volume of the paperweight?

we know that

The volume of the paperweight is equal to the volume of the square pyramid plus the volume of the cube

step 1

Find the volume of the pyramid

The volume of the pyramid is equal to

$V=\frac{1}{3}BH$

where

B is the area of the square base

H is the height of the pyramid

$B=8^{2}=64\ cm^{2}$

$H=3\ cm$

substitute

$V=\frac{1}{3}(64)(3)=64\ cm^{3}$

step 2

Find the volume of the cube

The volume of the cube is equal to

$V=b^{3}$

$V=8^{3}=512\ cm^{3}$

step 3

Find the volume of the paperweight

$64\ cm^{3}+512\ cm^{3}=576\ cm^{3}$

Part 2) what is the total surface area of the paperweight?

we know that

The total surface area of the paperweight is equal to the surface area of 5 faces of the cube plus the lateral area of the pyramid

step 1

Find the surface area of 5 faces of the cube

$SA=5b^{2}$

$SA=5(8^{2})=320\ cm^{2}$

step 2

Find the lateral area of the pyramid

$LA=4[\frac{1}{2}bh]$

$LA=4[\frac{1}{2}(8)(5)]=80\ cm^{2}$

step 3

Find the total surface area of the paperweight

$320\ cm^{2}+80\ cm^{2}=400\ cm^{2}$

8 0

3 years ago

Seven years ago, kodi found a box of old baseball cards in the garage. since then, he has added a consistent number of cards to

Katena32 [7]

(3,52)(7,108)
slope = (108 - 52) / (7 - 3) = 56/4 = 14

y = mx + b
slope(m) = 14
(3,52)...x = 3 and y = 52
sub and find b, the y int (the original amount of cards)
52 = 3(14) + b
52 = 42 + b
52 - 42 = b
10 = b

so ur equation is y = 14x + 10....with x being the number of years and y being the total cards. <== ur equation is y = 14x + 10

He started with 10 cards....and has been adding 14 cards every year.

so after 10 years...
y = 14(10) + 10
y = 140 + 10
y = 150 <== after 10 years, he will have 150 cards

8 0

3 years ago

Find the area of the triangle formed by the origin and the points of intersection of parabolas y=− 1/3 (x−1)^2+8 and y=x^2−2x−3.

Kobotan [32]

Answer:

The area of the triangle formed by origin, and the points (4,5) and (-2,5) will be 15 sq. units.

Step-by-step explanation:

The two parabolas are $y = - \frac{1}{3}(x - 1)^{2} + 8$ and y = x² - 2x - 3.

Now, solving those two equations we will get the points of intersection.

So, $- \frac{1}{3}(x - 1)^{2} + 8 = x^{2} - 2x - 3$

⇒ - (x - 1)² + 24 = 3x² - 6x - 9

⇒ -x² + 2x - 1 + 24 = 3x² - 6x - 9

⇒ 4x² - 8x - 32 = 0

⇒ x² - 2x - 8 = 0

⇒ x² - 4x + 2x - 8 = 0

⇒ (x - 4)(x + 2) = 0

So, x = 4 or - 2.

Now, for x = 4 , y = 4² - 2(4) - 3 = 5 and the point of intersection is (4,5).

Or, for x = - 2, y = (- 2)² - 2(- 2) - 3 = 5 and the point of intersection is (-2,5).

Now, points (4,5) and (-2,5) make a straight line parallel to the x-axis at a perpendicular distance of 5 units from origin and its length is (4 - (- 2)) = 6 units.

So, the area of the triangle formed by origin, and the points (4,5) and (-2,5) will be = 0.5 × 5 × 6 = 15 sq. units. (Answer)

8 0

3 years ago

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the bui

Sedbober [7]

Answer:

0.675 m/s

Step-by-step explanation:

Let height of shadow= y,CD=x

Height of man=2 m

Speed of man= $\frac{dx}{dt}=1. 8 m/s$