All categories

3 years ago

Are the points B, F, and W coplanar?

Mathematics

2 answers:

tankabanditka [31]3 years ago

5 0

Show the graph I can’t answer without the graph

Alla [95]3 years ago

4 0

Where’s the picture of the angle?

You might be interested in

2/3 of a pizza has olives on it . If Vivian eats 1/2 of the olive pizza , what fraction of the whole pizza did she eat ? Comment

My name is Ann [436]

Answer:

33% of a whole pizza

Step-by-step explanation:

2/3 of 100% = 66%

so 66% of the pizza has olives

she eat the half of that and the half from 66% is 33%

8 0

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

Please help me out with this Algebra 1 question

Zarrin [17]

Answer:

linear

Step-by-step explanation:

it would be linear because there is a constant rate of change

3 0

3 years ago

Read 2 more answers

Algebra 2 help please

suter [353]

Blank 1: 3

Blank 2: 27

Blank 3: 21

you have to distribute the three to everything inside the parenthesis, and because it's directly in front of the parenthesis it's being multiplied.

3(x - 9) = 3x - 27

Now you need to isolate the y, in this e ample you need to add 6 to both sides

the final equation should be:

y = 3x - 21

7 0

2 years ago

Solve for the value of x.<br> ABC<br> 20<br> 28<br> 9x-8

Marizza181 [45]

28+20+9x-8=180

48+9x-8=180

9x=180-40

9x=140

×=15.56

6 0

2 years ago