3 years ago

) 248 students went on a field trip. Six

Mathematics

1 answer:

mezya [45]3 years ago

7 0

Answer:

<h3>38 students were on each bus</h3>

Step-by-step explanation:

248 - 20 = 228

228 ÷ 6 = 38

You might be interested in

I WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Aleonysh [2.5K]

The answer would be D yeahhhhhh

3 0

3 years ago

Read 2 more answers

Need help and I need to show my work ?? Can someone help

vodka [1.7K]

Answer is D.
Just plug in 2 for x and -2 for y. And solve

7 0

3 years ago

Which expression is equivalent to...? Screenshots attached, please help.

BaLLatris [955]

For this case we must simplify the following expression:

$\sqrt [3] {64 * a ^ 6 * b ^ 7 * c ^ 9}$

We rewrite:

$64 = 4 ^ 3\\a ^ 6 = (a ^ 2) ^ 3\\b ^ 6 * b = ((b ^ 2) ^ 3 * b)\\c ^ 9 = (c ^ 3) ^ 3$

So:

$\sqrt [3] {4 ^ 3 * (a ^ 2) ^ 3 * (b ^ 2) ^ 3 * b * (c ^ 3) ^ 3} =\\\sqrt [3] {4 * a ^ 2 * b ^ 2 * c ^ 3) ^ 3 * b} =$

By definition of properties of powers and roots we have:

$\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}$

So:

$4a ^ 2b ^ 2 c ^ 3 \sqrt [3] {b}$

Answer:

Option B

3 0

3 years ago

H(x)=1/8x^3-x^2<br><br> Over which interval does h have a positive average rate of change?

lara31 [8.8K]

Answer:

$(-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x\frac{16}{3}$

Step-by-step explanation:

$h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \ respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x$

For positive rate of change $h'(x)>0$

$\frac{3}{8} x^{2} -2x>0\\x(\frac{3}{8}x-2)>0\\\\ When \ x0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x\frac{16}{3} \\\\$

6 0

4 years ago

Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Show the correct substitution of the values

Grace [21]

I hope this helps you

5 0

3 years ago