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2 years ago

8

The students in Mr. Nolan's class are writing expressions for the perimeter of a rectangle of side length ? and width w.

1 answer:

uranmaximum [27]2 years ago

6 0

You wanna know witch student is correct?

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Solve the system:<br><br><br> x+4y=1<br><br><br> 3x+5y=10

never [62]

Answer:

$x=6.25\\y=-1.25$

Step-by-step explanation:

$3x+5y=10\\x+3y=1\\\\$

Second equation times 3:

$3x+9y=3$

1st equation minus equation above:

$-4y=7\\y=-7/4$

Insert to equation:

$3x+\frac{-35}{4}=10\\12x-35=40\\12x=75\\x=75/12=6.25$

4 0

2 years ago

PLEASE HELP ME! I am stuck on 12 and 13. Thanks in advanced! :)

DiKsa [7]

Answer: 12. 75

13. 105

5 0

2 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

2 years ago

The height of a triangle is 7 cm longer than its base.

katrin [286]

A) Height -7 = base
B) .5 * height * base = 60
Substituting A into B
B) .5 * height * (height -7) = 60
B) .5*height^2 -3.5*height = 60
B) .5*height^2 -3.5*height -60 = 0
Using the quadratic formula:
Height = 15
Subtracting 7 gives us the base length
Base = 8
**************************************************************
Double-Check
.5 * 15 * 8 = 60

6 0

2 years ago

David bought a poster for an art project. The poster is 2.7 ft wide and 3.9 ft tall. What is the area of the poster? Enter your

algol13

10.53 ft is the Area. :D:D

7 0

2 years ago