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yawa3891 [41]
3 years ago
13

The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 5 cm/s. when the length is

14 cm and the width is 9 cm, how fast is the area of the rectangle increasing?
Mathematics
1 answer:
Zepler [3.9K]3 years ago
8 0
To be honest i really dont know sorry
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Vlad [161]

Use the formula a^(x/n) = (n)√a^x   (note it is a small n)

(5x^4y^3)^(2/9) = Small 9

Convert.

\sqrt{9{{(25x^{8})y^9 }}} is your answer

hope this helps

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3 years ago
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OwOoWoOwO pls answer, sorry about the random burst im really bad at math
Nataly [62]

Answer:

0.46

Step-by-step explanation:

42.78/m = 93

=> m = 42.78/93

=> m = 0.46

Hope my answer helps :)

7 0
2 years ago
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PLEASEE HELP ME
allsm [11]
A. It says “If Nancy is having a birthday then joe will buy her a card” it could Also be C. But it says “if joe buys Nancy a card he will GIVE IT TO HER” not just buying it so the most likely answer is A
4 0
2 years ago
Yuto and Riko went for a bike ride on the same path. When Riko left their house, Yuto was 5.25 miles along the path. If Yuto’s a
Nimfa-mama [501]

Answer:

Step-by-step explanation:

When Riko left his house, Yuto was 5.25 miles along the path.

Average speed of Riko = 0.35 miles per minute

Average speed of Yuto = 0.25 miles per minute

First we will calculate the time in which Riko will catch Yuto on the track.

Relative velocity of Riko as compared to Yuto will be = velocity of Riko - velocity of Yuto

= 0.35 - 0.25

= 0.10 miles per minute

Now we this relative velocity tells that Riko is moving and Yuto is in static position.

By the formula,

Average speed = \frac{Distance}{Time}

0.10 = \frac{5.25}{t}

t = \frac{5.25}{0.1}

t = 52.5 minutes

Now we know that Rico will catch Yuto in 52.5 minutes. Before this time he will be behind Yuto.

So duration of time in which Rico is behind Yuto will be 0 ≤ t ≤ 52.5

Here time can not be less than zero because time can not be with negative notation. It will always start from 0.

3 0
3 years ago
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Find the integral using substitution or a formula.
Nadusha1986 [10]
\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx

Derivative of the denominator:
\rm (x^2+2x+5)'=2x+2

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

Based on our previous test, we know that a simple substitution will work for the first integral: \rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx

So the first integral changes,

\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

integrating to a log,

\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2

So we have this going on,

\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx

Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du

simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
and include a constant of integration,

\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c

Hope that helps!
Lemme know if any steps were too confusing.

8 0
3 years ago
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