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3 years ago

2,400 in.=_ft. Thanks guys!!!!!

Mathematics

2 answers:

Flauer [41]3 years ago

8 0

Answer:

2,400in equals 200ft

Step-by-step explanation:

sasho [114]3 years ago

6 0

200 ft...............

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Svetradugi [14.3K]

You can use the substitution method for this problem
Since -3y+8=x you can
Use the equation -7(-3y+8) -6y=4
You then distribute and solve
21y-56-6y=4
15y=60
Y=4
You then plug y into one of the original equations
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3 years ago

Covert to find the equivalent rate

Mice21 [21]

Answer:

Step-by-step explanation: you have to change day into 24 and hour into 1 since there are 24 hours in a day so once you do that if you divide the left equation by 24 to match the denominators then you have to divide the numerator(744) by 24 as well to get pounds on the right side which comes out to be 31

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3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

3 years ago

What is 2/3 + 1/4<br>*please help*

Tcecarenko [31]

$\frac{2}{3}+\frac{1}{4}$

Re-write with a common denominator, in this case 12:

$\frac{2 \times 4}{3 \times 4}+\frac{1 \times 3}{4 \times 3}$
$\frac{8}{12}+\frac{3}{12}$

Add:
$\frac{8}{12}+\frac{3}{12} = \frac{11}{12}$

6 0

3 years ago

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Can someone please help me with this

kvv77 [185]

Answer:no not in amillion years

Step-by-step explanation:hehe nahhhhhhhhhh lll

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3 years ago