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Tasya [4]
3 years ago
7

If m = 5 and n = -2, evaluate 3m - 4n and n2 - m and find the product of the two expressions.

Mathematics
1 answer:
ddd [48]3 years ago
8 0

Hence the correct answer is C) -23

Further explanation:

We have to put the values of m and n into both expression first separately and then multiply the answers.

Given expressions are:

Exp\ 1:3m-4n\\Exp\ 2: n^2-m

The values are:

m=5\\n=-2

Putting the values in expression 1:

3m-4n\\=3(5)-4(-2)\\=15+8\\=23

Putting the values of m and n in second expression

n^2-m\\=(-2)^2-5\\=4-5\\=-1

Multiplying both expressions:

(3m-4n)(n^2-m)=23* (-1)=-23

Hence the correct answer is C) -23

Keywords: Polynomials, Putting values in expressions

Learn more about polynomials at:

  • brainly.com/question/4142886
  • brainly.com/question/4934417

#LearnwithBrainly

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Kyle's height increased 17 % last year and 12% this year. What is the total percent increase ?
Advocard [28]

Answer:

31.04%

Step-by-step explanation:

So he increased 17%, then increased 12%.

When he increased by 17%, his height was 117%, relative to his former height.  12% of 117% is 14.04%

So he increase 17%, then increased 14.04% ( of his original height) so his total growth was 31.04%

6 0
3 years ago
Maria's math test had 25 questions she got 84% correct how many did she get wrong
inysia [295]
\frac{x}{25}=84% \\  \frac{x}{25}=.84 \\ 25( \frac{x}{25})=(.84)(25) \\ x=21
Maria got 21 correct.

Number wrong = Total - Number right
y = 25 - 21
y = 4

Maria got 4 wrong.

Hope this helps!

7 0
3 years ago
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$90.60/12 2/5 = $22.60 /c
GarryVolchara [31]
The awnser is c=7.48
7 0
3 years ago
Which statement is true? WILL GIVE BRAINLIEST! (If you are right!)
Setler79 [48]
I would try B.

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4 0
3 years ago
Read 2 more answers
The region bounded by y=(3x)^(1/2), y=3x-6, y=0
Ganezh [65]

Answer:

4.5 sq. units.

Step-by-step explanation:

The given curve is y = (3x)^{\frac{1}{2} }

⇒ y^{2} = 3x ...... (1)

This curve passes through (0,0) point.

Now, the straight line is y = 3x - 6 ....... (2)

Now, solving (1) and (2) we get,

y^{2} - y - 6 = 0

⇒ (y - 3)(y + 2) = 0

⇒ y = 3 or y = -2

We will consider y = 3.

Now, y = 3x - 6 has zero at x = 2.

Therefor, the required are = \int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx

= \sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}

= [\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} }  }{3}] - [13.5 - 18 - 6 + 12]

= 6 - 1.5

= 4.5 sq. units. (Answer)

7 0
3 years ago
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