1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
MA_775_DIABLO [31]
3 years ago
6

Help me pls its easy

Mathematics
2 answers:
Likurg_2 [28]3 years ago
6 0
Pretty sure you do 6,300/70 to get 90
valentina_108 [34]3 years ago
5 0

Answer:

6300/70=90

Step-by-step explanation:

division!!

You might be interested in
How do I solve this ? Help ASAP!!! Steps please
elena-14-01-66 [18.8K]

Answer:

Step-by-step explanation:

start by gathering all the x together.

6x-3x-2x=1x

what you have left is 1x+4=?x+?

3 0
4 years ago
Read 2 more answers
Write the equation representing this line slope -4 and point (-2,5)
Flura [38]
5=(-4)x-2. Hope this helps
8 0
3 years ago
Find the values of c such that the area of the region bounded by the parabolas
evablogger [386]
For there to be a region bounded by the two parabolas, you first need to find some conditions on c. The two parabolas must intersect each other twice, so you need two solutions to

16x^2-c^2=c^2-16x^2

You have

32x^2=2c^2\implies 16x^2=c^2\implies x=\pm\dfrac{|c|}4

which means you only need to require that c\neq0. With that, the area of any such bounded region would be given by the integral

\displaystyle\int_{-4|c|}^{4|c|}\bigg((c^2-16x^2)-(16x^2-c^2)\bigg)\,\mathrm dx

since c^2-16x^2>16x^2-c^2 for all c\neq0. Now,

\displaystyle\int_{-|c|/4}^{|c|/4}\bigg((c^2-16x^2)-(16x^2-c^2)\bigg)\,\mathrm dx=2\int_0^{|c|/4}(2c^2-32x^2)\,\mathrm dx

by symmetry across the y-axis. Integrating yields

\displaystyle2\int_0^{|c|/4}(2c^2-32x^2)\,\mathrm dx=4\int_0^{|c|/4}(c^2-16x^2)\,\mathrm dx
=4\left[c^2x-\dfrac{16}3x^3\right]_{x=0}^{x=|c|/4}
=c^2|c|-\dfrac{|c|^3}3
=\dfrac{2|c|c^2}3=144
|c|c^2=216

Since 216=6^3, you have c=\pm6.
3 0
3 years ago
What is the simplest form of ^4 sqrt 81x^8y^5
lara31 [8.8K]

ANSWER

3 {x}^{2} y \sqrt[4]{y}

EXPLANATION

We want to simplify:

\sqrt[4]{81 {x}^{8} {y}^{5}  }

We can split the radical sign to obtain:

\sqrt[4]{81}  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{5} }

Or

\sqrt[4]{81}  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{4}  \times y}

\sqrt[4]{81}  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{4}}  \times  \sqrt[4]{y}

\sqrt[4]{ {3}^{4} }  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{4}}  \times  \sqrt[4]{y}

Recall that:

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

{3}^{4 \times  \frac{1}{4} }  \times  {x}^{8 \times  \frac{1}{4}  }  \times  {y}^{4 \times  \frac{1}{4} }\times  \sqrt[4]{y}

3 {x}^{2} y \sqrt[4]{y}

6 0
4 years ago
Read 2 more answers
What is the minimum width of the square base?
riadik2000 [5.3K]
4 I think I might be wrong so sorry if I am wrong
4 0
4 years ago
Other questions:
  • Solve the mathematical puzzle. Determine the digits of F from these clues. The digits of F are all the same. The sum of all the
    11·2 answers
  • Tom jogged 3/5 mile on Monday and 2/6 mile on Tuesday. How much farther did Tom jog on Monday than on Tuesday?
    8·1 answer
  • True or false, 100% is larger than 1​
    10·1 answer
  • Determine consecutive integer values of x between which each real zero is located.
    13·1 answer
  • A power plant has 4 tons of coal. A ton of coal produces 2460 kilowatt hours of electricity. Is this enough to power 9 light bul
    5·1 answer
  • 11k = -143 I need help with this question
    6·2 answers
  • You and Mr. Fowles have decided to race each other at the Boston Marathon (26 miles long). Mr. Fowles runs at a steady 4.2 miles
    12·1 answer
  • What does x equal in 150 = -6(x)^2 + 100x -180
    12·2 answers
  • Tyrell has a bankruptcy on his credit report and therefore pays higher interest rates on his current loans. he calculates that t
    15·1 answer
  • Someone help me!!!<br> please...
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!