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3 years ago

Can someone please help me with math.

Mathematics

1 answer:

pychu [463]3 years ago

4 0

Answer:

Step-by-step explanation:

That is a negative correlation. As x increases, y decreases. Notice that as you read from left to right, the points are going down which means negative.

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The temperature fell from 0 degrees to 6 3/5 degrees below 0 in 2 1/5 hours. What was the temperature change per hour?

cestrela7 [59]

Answer: -3 degrees per hour.

Step-by-step explanation:

Given: The temperature fell from 0 degrees to $6\frac{3}{5}$ degrees below 0 in $2\frac{1}{5}$ hours.

The change in temperature= $0-6\frac{3}{5}$

$=-\frac{33}{5}$ degrees

Time taken to change the temperature= $2\frac{1}{5}=\frac{11}{5}$ hours

Now, the temperature change per hour is given by :-

$\frac{\text{Change in temperature}}{\text{Time}}=\frac{-\frac{33}{5}}{\frac{11}{5}}\\\\=\frac{-33\times5}{11\times5}=-3\text{ degrees per hour}$

7 0

3 years ago

Read 2 more answers

Rebekka had two lengths of string, 3 1/10 inches and 4 8/9 inches. About how many inches did she

kondor19780726 [428]

Answer:

Rebekka has about 10 inches of string. (The exact number is 7.988888889 inches of string)

Step-by-step explanation:

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3 years ago

A bacteria colony grow at a rate of 11% per year. If the initial sample has a mass of 500 grams,

Ber [7]

A hope it helps u Alabama s

5 0

3 years ago

[ HELP ASAP ]

butalik [34]

Answer:

look below

Step-by-step explanation:

4 0

3 years ago

Find all critical numbers of the function <img src="https://tex.z-dn.net/?f=g%28x%29%3Dx%5E%7B%5Cfrac%7B3%7D%7B4%7D%20%7D-2x%5E%

postnew [5]

Differentiate g using the power rule:

$g'(x) = \dfrac34 x^{\frac34-1} - 2\cdot\dfrac14 x^{\frac14-1}$

$g'(x) = \dfrac34 x^{-\frac14} - \dfrac12 x^{-\frac34}$

$g'(x) = \dfrac14 x^{-\frac34} \left(3 x^{\frac12} - 2\right)$

$g'(x) = \dfrac{3\sqrt x - 2}{4 x^{\frac34}}$

The critical points of g occur where g' is zero or undefined.

We have

$g'(x) = 0 \implies 3\sqrt x - 2 = 0 \implies \sqrt x = \dfrac23 \implies \boxed{x = \dfrac49}$

and the derivative is undefined for

$\dfrac1{g'(x)} = 0 \implies 4x^{\frac34} = 0 \implies \boxed{x=0}$

8 0

3 years ago