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alexira [117]
2 years ago
10

HELP !!! (will mark brainliest)

Mathematics
2 answers:
Aleksandr [31]2 years ago
5 0

Solution - Verifying Option A:

(6^{4})^{-5} < (6^{-7}) \times (6^{-3})\\ (6^{-20}) < (6^{-7-3} )\\ 6^{-20}< 6^{-10}

Result: <u>True (Option A is correct)</u>

Solution - Verifying Option B:

(6^{4})^{-5} > (6^{-7}) \times (6^{-3})\\ (6^{-20}) > (6^{-7-3} )\\ 6^{-20}> 6^{-10}

Result: <u>False (Option B is incorrect)</u>

Solution - Verifying Option C:

(6^{4})^{-5} = (6^{-7}) \times (6^{-3})\\ (6^{-20}) = (6^{-7-3} )\\ 6^{-20}= 6^{-10}

Result: <u>False (Option C is incorrect)</u>

Option A is correct.

Rasek [7]2 years ago
3 0

Answer:

I think it's A. (6^{4})^{-5}

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At a​ school, 174 students play at least one sport. This is 75​% of the students at the school. How many students are at the​ sc
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Answer:

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Step-by-step explanation:

Let s represent the number of students at the school. We are told ...

  174 = 75% × s

  174/0.75 = s = 232 . . . . . divide by the coefficient of s

There are 232 students at the school.

3 0
3 years ago
Expand (4x-3y)^4 using pascal's triangle ...?
jenyasd209 [6]
Using row 4: 

<span>coefficients are: 1, 4, 6, 4, 1 </span>

<span>a^4 + a^3b + a^2b^2 + ab^3 + b^4 </span>

<span>Now adding the coefficients: </span>

<span>1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4 </span>

<span>Substitute a and b: </span>

<span>a = 4x </span>
<span>
b = -3y </span>

<span>1(4x)^4 + 4(4x)^3(-3y) + 6(4x)^2(-3y)^2 + 4(4x)(-3y)^3 + 1(-3y)^4 </span>

<span>Now simplify the above: </span>

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8 0
3 years ago
Please help me ASAP
stellarik [79]

Answer:

<em>The Graph is shown below</em>

Step-by-step explanation:

<u>The Graph of a Function</u>

Given the function:

\displaystyle y=g(x)=-\frac{3}{2}(x-2)^2

It's required to plot the graph of g(x). Let's give x some values:

x={-2,0,2,4,6}

And calculate the values of y:

\displaystyle y=g(-2)=-\frac{3}{2}(-2-2)^2=-\frac{3}{2}(-4)^2==-\frac{3}{2}*16=-24

Point (-2,-24)

\displaystyle y=g(0)=-\frac{3}{2}(0-2)^2=-\frac{3}{2}(-2)^2=-\frac{3}{2}*4=-6

Point (0,-6)

\displaystyle y=g(2)=-\frac{3}{2}(2-2)^2=-\frac{3}{2}(0)^2=0

Point (2,0)

\displaystyle y=g(4)=-\frac{3}{2}(4-2)^2=-\frac{3}{2}(2)^2=-\frac{3}{2}*4=-6

Point (4,-6)

\displaystyle y=g(6)=-\frac{3}{2}(6-2)^2=-\frac{3}{2}(4)^2=-\frac{3}{2}*16=-24

Point (6,-24)

The graph is shown in the image below

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2 years ago
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Marat540 [252]
W - a width
3w - a length
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Therefore:

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The perimeter:

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3 years ago
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valentina_108 [34]
Option D
I hope that is right
3 0
2 years ago
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