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docker41 [41]
2 years ago
6

Can someone help me with this please? I don't know how to do it, and I need the answer asap ;-; (Image Below)

Mathematics
2 answers:
Ostrovityanka [42]2 years ago
7 0
There isn’t a picture
Marizza181 [45]2 years ago
4 0

Answer:

ma'am or sir... there is no picture attached... but whats the question?

Step-by-step explanation:

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X^2 =80. Write the exact answer using radicals
DanielleElmas [232]
Answer:

4√5, -4√5

Step-by-step explanation:

x^2 = 80

√x^2 = √80

x = 8.9 or 4√5
8 0
3 years ago
A person who is 2 m tall casts a shadow that is 5 m long. At the same time, a building casts a shadow that is
VMariaS [17]

Answer:

9.6

Step-by-step explanation:

First you figure out that the shadow is 2.5 times larger than what is casting it. Then you divide 24 by 2.5.

4 0
2 years ago
Study the function below. f(x) = –3x Label the function as proportional or non-proportional. Explain your reasoning.
Nimfa-mama [501]
Y = f(x) = -3x y = -3x y = (constant) x y is proportional to x.<span> Hence we can say Y is proportional to x , having (-3) as proportionality Constant. hope this helps</span>
3 0
3 years ago
PLZZZ HELP ASAP TYSM I NEED THIS ASAP
uysha [10]

Explanation:

See answer for explanation.

Answer:

5x-3+2x=x+7+6x

Answer: There are no solutions.

I'm pretty sure this is correct, sorry if it's not.

Have a lovely evening!

3 0
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
2 years ago
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