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levacccp [35]
4 years ago
9

Find the value of 2a^2+5b^2 when a=-6 and b=2

Mathematics
2 answers:
Margaret [11]4 years ago
5 0

2(6)^2+5(2)^2

2*36+5(2)^2

2*36+5*4

72+5*4

72+20

=92

frutty [35]4 years ago
3 0
HEYA!

here is your answer:

2a^2+5b^2

a = 6 & b= 2

2( { - 6)}^{2}  + 5( {2)}^{2}

➡ 2 (36) + 5(4)

➡ 72 + 20

➡ 92
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2n+(7n+8)= what is the missing coefficient
Taya2010 [7]

Assuming that you were to simplify this, you'd get 2n + 7n + 8 = 9n + 8.

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Which ratio is larger, 10 to 15 or 14 to 21?
m_a_m_a [10]

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I am pretty sure that is it 14 to 21

Step-by-step explanation:

Sorry if i am wrong

7 0
3 years ago
A $300 suit is marked down by 20%. Find the sale price rounded to the nearest dollar.
Debora [2.8K]

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240 is the mark down.

Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
Write the system of linear equations <br> Y=3x+1<br> Y=2x+3
Firlakuza [10]

Answer:

x = 2

y =7

Step-by-step explanation:

the system of linear equation will be solve using substitution method,

let

Y=3x+1.............................................. equation 1

Y=2x+3.............................................. equation 2

substitute equation 1 into equation 2

3x + 1 = 2x + 3

combine the like terms

3x+1 -2x = 3

x = 3 -1

x = 2

put the value of x=2 in either equation 1 or 2

Y=2x+3.............................................. equation 2

y = 2(2) + 3

y = 4 + 3

y = 7

7 0
3 years ago
If a right triangle's hypotenuse is 17 units long, and one of its legs is 15 units long, how long is the other leg?
Vesna [10]

Answer:

8 units

Step-by-step explanation:

Hello!

So, there's a formula we can apply to right-angled triangles: Pythagorean's theorem. It states that  c = \sqrt{{a}^2 + b^{2} }, where <em>c</em> is the hypotenuse and <em>a</em> and <em>b </em>are the legs of the triangle.

So, from the problem,  if <em>c </em>= 17 and <em>a </em> = 15, then, we're solving for <em>b</em>. So we'll rewrite the theorem to solve for <em>b</em>.

{c}^2 = {a}^2+{b}^2\\{c}^2-{a}^2={b}^2\\{b} = \sqrt{{c}^2-{a}^2}

Okay, so now we have isolated the theorem for <em>b. Let's </em>plug in our values for <em>c </em>and <em>a</em>.

b = \sqrt{{17}^2-{15}^2}\\b = \sqrt{289-225}\\b = \sqrt{64}\\b = 8

So, using the theorem, we found <em>b</em> = 8. To check our work, let's plug in <em>b</em> and <em>a</em> and solve for <em>c.</em>

<em />c = \sqrt{{a}^2+{b}^2}}\\c = \sqrt{{15}^2+{8}^2}\\c = \sqrt{225+64}\\c = \sqrt{289}\\c = 17\\<em />

So, we got our hypotenuse to equal 17 units, which is correct! So, our <em>b</em> is correct too. Awesome

7 0
3 years ago
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