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3 years ago

7

When Tim says that word problems don't use mathematical notation, what does he mean

1 answer:

Zepler [3.9K]3 years ago

5 0

Answer:they dont use symbols such as x = + - / % or *

Step-by-step explanation: just look up what mathematical notation means dude

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Need help on number 4

Vaselesa [24]

<span>There are 2 3/4 boxes and 1/2 pies in each box how many pies are there? </span>

8 0

3 years ago

How much will a person pay for 7.5 pounds of bananas at a price of $2.07 per pound?

igomit [66]

Answer:

$15.57

Step-by-step explanation:

2.07 times 7.50 = $15.57!

7 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

Which graph represents the equation x^2 = 8y?

nirvana33 [79]

2 graph I think not certain

7 0

2 years ago

Read 2 more answers

. help needed please

Anastasy [175]

In order to do this, we would use order of operations. First, Parentheses, which is division with fractions, so we would do KCF,keep it, change it, flip it. keep the first fraction, change the operation to multiplication, then flip the next fraction, making it -7/9 x 3/1. the answer would be -2.33. multiply that by -2, and you would get -4.66, which is your answer. PLEASE MARK BRAINLIEST IF HELPFUL! :))

5 0

3 years ago