Jose wants to build a model of a 18-meter tall building. He will be using a scale

How many cubic centimeters is 5 cubic meters.

tresset_1 [31]

The key to solving this question is first knowing how many cubic cm are in one cubic m:

The ratio is always 1000000 cubic cm in 1 cubic meter.

Knowing this, we calculate how many cubic cm are in 5 cubic meters!

1000000 cubic cm = 1 cubic m
x cubic cm 5 cubic m

x = 1000000 cubic cm x 5 cubic m ÷ 1 cubic m
x = 5000000 cubic cm

Hope this helps!
Feel free to message me if you have any questions :)

4 0

2 years ago

Read 2 more answers

A monomial of the 2nd degree with a leading coefficient of 3

Rzqust [24]

Given:

Polynomials

To find:

Monomial of 2nd degree with leading coefficient 3

Solution:

Monomial is an algebraic expression with only one term.

Option A: $3 n^{2}-1$

It is not a monomial because it have 2 terms.

It is not true.

Option B: $3 n-n^{2}$

It is not a monomial because it have 2 terms.

It is not true.

Option C: $3 n^{2}$

It have one term only. So, it is a monomial.

Degree means highest power. So degree = 2

Leading coefficient means the value before variable.

Leading coefficient = 3

It is true.

Option D: $2n^3$

It have one term only. So, it is a monomial.

Degree means highest power. So degree = 3

It is not true.

Therefore $3 n^{2}$ is a monomial of 2nd degree with a leading coefficient of 3.

5 0

2 years ago

Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the

Rama09 [41]

Question:

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s 2s 3s 4s

Answer:

It takes 2 seconds for object to hit the ground

Solution:

The given equation is:

$h(t) = -16t^2 + v_0t+h_0$

Initial velocity = 27 feet/sec

$h_0 = 10\ feet$

Therefore,

$h(t) = -16t^2 +27t+10$

At the point the object hits the ground, h(t) = 0

$-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0$

Solve by quadratic formula,

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125$