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denpristay [2]
3 years ago
6

Katie waters the garden every 3 days and weeds it every 4 days. She does both on April 2nd. What is the next date that she will

water and weed the garden?
Mathematics
1 answer:
GaryK [48]3 years ago
3 0

Answer:

So every 12 days she waters and weeds.

It will happen again on April 14

Step-by-step explanation:

List multiples of 4: 4x1 = 4, 4x2=8, 4x3=12, 4x4=16, 4x5=20

Now which one of these multiples divides evenly by 3?


Hope this helps!


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120% of a number is 45 find the number
Bezzdna [24]
Hello,

Here is your answer:

The proper answer to this question is "37.5".

Your answer is 37.5!

If you need anymore help feel free to ask me!

Hope this helps!
3 0
3 years ago
A cylinder has a height of 10 inches and a radius of 17 inches. What is its volume? Use ​ ≈ 3.14 and round your answer to the ne
Gelneren [198K]

Steps:

Radius= 17in

Height= 10in

Formula: V=πr2h

Shape: Right cylinder

Solved for volume

H= Height

R= Radius

Description:

We know that R is 17 inches, and H is 10 inches. Since we are finding the volume we use the formula V=πr2h to calculate. After calculating your answer will come as V≈9079.2in³. You also said to round your answer to the nearest hundredth. After rounding your answer to th enearest hundredth your answer will come as 9,079.20.

Answer: V≈9079.2in³

Round to the nearest hundredth = 9,079.20

Please mark brainliest

<em><u>Hope this helps.</u></em>

8 0
3 years ago
7n - 4 = 31. what is the value of the n​
Gelneren [198K]

Answer:

n = 5

Step-by-step explanation:

- Move all terms not containing  n  to the right side of the equation.

7n = 31 + 4

7n = 35

- Divide each term by  7  and simplify.

n = 5

i hope this helps :)

4 0
3 years ago
Read 2 more answers
What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
Solve for W -5.1=1.3+w/4
Lera25 [3.4K]
-5.1=1.3+w/4
Subtract 1.3 from both sides
-6.4=w/4
Multiply both sides by 4
-25.6=w

Check work:
-5.1=1.3+(-25.6/4)
-5.1=1.3+(-6.4)
-5.1=1.3-6.4
-5.1=-5.1

Hope this helps!  :)
7 0
3 years ago
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