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3 years ago

7

Hmark 1 - Item 205023

1 answer:

Softa [21]3 years ago

3 0

Answer:

82% is her highest score

Step-by-step explanation:

82 is greater than 23, 80, 17 and 20

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Sylvanite is a mineral that contains 28.0% gold by mass. Determine how much sylvanite would you need to dig up to obtain 81.0 g

lidiya [134]

Since Sylvanite only contains 28% gold by mass, then to obtain 81.0 grams of gold, x amounts of sylvanite must be mined. To determine x, we divide 81 grams by 28%. This equation is shown below:

0.28X = 81

X = 81/0.28
X = 289.29 grams

Therefore, 289.29 grams of sylvanite must be dug up to obtain 81 grams of gold.

3 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

Can u pls help I don’t understand I’ll give u 15 points

Juliette [100K]

Answer: $\frac{4}{3}$

Step-by-step explanation:

This is a multiplication problem. You are multiplying $\frac{1}{3}$ by 4. This also means 4 divided by 3. They are both the same.

4 0

3 years ago

I need help on the Math question?

mrs_skeptik [129]

Answer:

The answer is D.

Step-by-step explanation:

3x*3x = 9x^2

3x*2= 6x

5*3x=15x

5*2=10

6x+15x=21x

9x^2+21x+10

3 0

3 years ago

(49p2–490p) ÷(p–10)<br><br> please help!!

horsena [70]

Answer:

(49p2–490p) ÷(p–10)

Step-by-step explanation:

8 0

2 years ago