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Yuliya22 [10]
2 years ago
9

Can someone help me?

Mathematics
2 answers:
Marina86 [1]2 years ago
8 0
M = -3, hope it was helpful! :)
Pani-rosa [81]2 years ago
5 0

Answer: m=−3

Step-by-step explanation: −40−2(3m+1/2)=7m−2

−40+(−2)(3m)+(−2)(1/2)=7m+−2

−40+−6m+−1=7m+−2

(−6m)+(−40+−1)=7m−2

−6m+−41=7m−2

−6m−41−7m=7m−2−7m

−13m−41+41=−2+41

−13m/−13=39/−13

m=−3

What mistake I guess Keith did make is he subtracted 2 from -39 which equaled to -37 which caused him divide -37 by 13 when it should have been 39 divided by 13 because he should have left 39 alone and not have subtracted 2 from it also it should not have been negative basically what I'm trying to say is that he did his division and subtraction wrong.

 

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3n + 2 \geqslant 8

Step-by-step explanation:

2 more than 3 time mean 3n+2

"is atleast 8" means it can be 8 or more so c. is right answer

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3 years ago
Simplify (x^4-4x^3+9x^2-5x) (x^3-3x^2+7) -30x^2+25x^4
Masja [62]

Answer:

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Step-by-step explanation:

there you go!

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3 years ago
An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i
Elanso [62]

Answer:

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim = \sqrt{(15-x)^2+1^2}

The total time, 't', is given as follows;

t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}

The minimum value of 't' is found by differentiating with an online tool, as follows;

\dfrac{dt}{dx}  = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} =  \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }

At the maximum/minimum point, we have;

\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89

Therefore, the distance to run, x ≈ 14.96 mile

6 0
3 years ago
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liq [111]

Answer:

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Step-by-step explanation:

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2 years ago
What’s the correct answer for this?
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Answer:

Step-by-step explanation:

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