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makvit [3.9K]
3 years ago
14

Factor 48 24(2m-1) 3(2m-1)^2

Mathematics
1 answer:
nataly862011 [7]3 years ago
6 0
4824(2m-1)x3(2m-1)² 
= 9648m-4824x12m-3
=115776m-4827
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Please answer with clear instructions so that i can apply this to other questions.
kvv77 [185]

Answer:

10 terms

Step-by-step explanation:

equate the sum formula to 55 and solve for n

\frac{1}{2} n(n + 1) = 55 ( multiply both sides by 2 to clear the fraction )

n(n + 1) = 110 ← distribute parenthesis on left side

n² + n = 110 ( subtract 110 from both sides )

n² + n - 110 = 0 ← in standard form

Consider the factors of the constant term (- 110) which sum to give the coefficient of the n- term (+ 1)

the factors are + 11 and - 10 , since

11 × - 10 = - 110 and 11 - 10 = + 1 , then

(n + 11)(n - 10) = 0 ← in factored form

equate each factor to zero and solve for n

n + 11 = 0 ⇒ n = - 11

n - 10 = 0 ⇒ n = 10

However, n > 0 , then n = 10

number of terms which sum to 55 is 10

7 0
2 years ago
Bill $42 , tax 9% , tip 18%
STatiana [176]

Answer:

Tax 3.78 $ and tip 7.56 $

Step-by-step explanation:

42*0.09 = 3.78

42*0.18 = 7.56

8 0
2 years ago
Solve for x round to the nearest tenth
lidiya [134]

Answer:

47.0

Step-by-step explanation:

In this right angle triangle, we are faced with a challenge of two sides. The opposite side and the adjacent side, hence the tangent is used.

Where it is the opposite side and hypothenus side, the sine is used and when it is the hypothenus side and adjacent side, the cosine is used.

Hence, we have tan62°=x/25

We cross multiply, to have

25(tan 62°)= x

x = 47.01816

In rounding up numbers, number 1 to 4 will be rounded up to zero, while numbers 5 to 9 will be rounded up to 1.

Rounding up 47.01816 to the nearest tenth. The tenth value is the figure is 0, before it we have 1, which is to hundredth. 1 will be rounded up to zero.

So we have 47.0

3 0
3 years ago
Help math question derivative!
atroni [7]
Let f(x)=\sec^{-1}x. Then \sec f(x)=x, and differentiating both sides with respect to x gives

(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1
f'(x)=\dfrac1{\sec f(x)\tan f(x)}

Now, when x=\sqrt2, you get

(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}

You have \sec^{-1}\sqrt2=\dfrac\pi4, so \sec\left(\sec^{-1}\sqrt2\right)=\sqrt2 and \tan\left(\sec^{-1}\sqrt2\right)=1. So (\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}
5 0
3 years ago
What is the perimeter of the figure
Fofino [41]

Answer:

Step-by-step explanation:

5 0
3 years ago
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