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telo118 [61]
3 years ago
15

Chiyo has a toolbox with the shape below. A rectangular prism with a length of 8 inches, width of 15 inches, and height of 6 inc

hes. A triangular prism with a triangular base with base 8 inches and height 3 inches. The prism has a height of 15 inches. Which statements are true about the toolbox? Select two options. The shape can be broken into a rectangular prism and a triangular prism. The shape can be broken into a rectangular prism and a rectangular pyramid. The formula for the volume of this figure is V = B h + one-third B h. The volume of the toolbox is 840 inches cubed The volume of the toolbox is 900 inches cubed
Mathematics
1 answer:
guajiro [1.7K]3 years ago
3 0

Answer:

Option A and E.

Step-by-step explanation:

It is given that Chiyo has a toolbox with the shape of a rectangular prism with a length of 8 inches, width of 15 inches, and height of 6 inches. A triangular prism with a triangular base with base 8 inches and height 3 inches. The prism has a height of 15 inches.  

The shape can be broken into a rectangular prism and a triangular prism.

So, option A is correct.

Volume of rectangular prism =l\times b\times h

Where, l is length , b is width and h is height.

V_1=8\times 15\times 6=720

Volume of triangular prism =Bh

Where, B is base area and h is height.

Area of triangular base is

B=\dfrac{1}{2}\times 8\times 3=12

So, volume of triangular prism is

V_2=12\times 15

V_2=180

So, volume of tool box is

V=V_1+V_2=720+180=900\text{ inches cubed}

Therefore, the correct option is E.

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The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9
mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

<em>The height </em>h(t)<em> of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation </em>h(t) = 3 +90\cdot t -16\cdot t^{2}<em>, where </em>t<em> is time in seconds. After how many seconds will the ball be 84 feet above the ground. </em>

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3+90\cdot t -16\cdot t^{2} = 84

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By Quadratic Formula:

t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}

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The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

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