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2 years ago

What is the volume of the cube I'm confused (look at the picture to see)

Mathematics

1 answer:

77julia77 [94]2 years ago

3 0

Answer: D (64)

Step-by-step explanation:

My answer is correct because, it's showing that 4 is the length, width, and height. So it bassically 4 x 4 x 4 in other words 4 exponent 3. And 4 exponent 3 would equal 64.

Hope This Helps!

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9^2= 9.5^2 + 4^2 - 2 (9.5) (4) (cosR) I'm trying to find out the angle measure of R

alexandr402 [8]

Sorry I can't solve it.

Firstly, the equation is wrong. It is supposed to be 81 ≠ 30.25

Second of all, this is the best I got 81= −76 (cos(r) + 106.25

5 0

3 years ago

The slope of a line is -1/4 and the y-intercept is 8. Whats the x-intercept?

Fed [463]

Slope = -1/4
y-interept = 8
⇒equation: y = -1/4x + 8

x-intercept ⇒ when y = 0
0 = -1/4x + 8
1/4x = 8
x = 8 x 4
x = 32

Answer: x-intercept = (32,0)

7 0

3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$