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3 years ago

Describe the cross-section of the rectangular prism

Mathematics

1 answer:

san4es73 [151]3 years ago

4 0

Answer:

The cross-section of the rectangular prism is a:

rectangle

Step-by-step explanation:

When you take a cross-section of a rectangular prism, regularly you will obtain a rectangle because this was the base or the two-dimension figure that was used to form the three-dimension figure, in this case, the rectangular prism, the form of obtaining other figure is using how reference the square face, in that case, the cross-section would be a square, this happens because the cross-section is bind with the reference face. Possibly you think the cross-section in the figure doesn´t appear a rectangle, this happens by the perspective because we are looking at the rectangular prism with an inclination of 45°, but if the inclination was 90°, you would see a blue rectangle.

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3 years ago

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3 years ago

10=c+d c is one more than d what is the value of c? with working out

Leya [2.2K]

C= 5.5 I'm pretty sure that's the answer

d= 4.5 and c+ 5.5
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since c is one more than d

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3 years ago

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daser333 [38]

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Your answer would be: On the third line, the student adds the 8 to both sides instead of subtracting. The way the initial equation is given is

y-(-8)=-6(x-2). After distributing the six, the student should make the 8 positive because subtracting a negative makes a positive. After solving, the equation should look like: y(+8)=-6x+12, so you would subtract the 8 from both sides instead of adding it, and solve from there.

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3 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago

Describe the cross-section of the rectangular prism￼

Describe the cross-section of the rectangular prism