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SOVA2 [1]
3 years ago
14

Which of the following inequalitties is correct

Mathematics
1 answer:
Airida [17]3 years ago
8 0

Answer:

You were right the correct option is

- a > - b

You might be interested in
The Pythagorean Theorem ONLY works on which triangle?​
Gekata [30.6K]

Answer:

right triangle

Step-by-step explanation:

that's the condition for the Pythagorean Theorem hold true

8 0
3 years ago
Read 2 more answers
Which equation represents the statement: Three times the sum of a number and 5 is –30?
castortr0y [4]

Answer:

3(n+5)=-30

Step-by-step explanation:

Three times the sum of a number and 5 is - 30

Vocabulary:

times = multiplication, represented with *

sum = addition, represented by +

unknown number = n

Three * the sum of a number and 5

The sum of a number and 5 is the same as saying n + 5

So we are multiplying 3 by n + 5

Because n is an unknown variable we have to separate the sum of n and 5 from being multiplied by 3 with parenthesis. We do this because if we want to multiply 3 by the sum of n and 5. If we put 3 * n + 5 we are only multiplying the unknown variable by 3 not including the 5 that is added to it.

So we get 3( n + 5 ) is -30

3(n+5)=-30 is the answer.

3 0
2 years ago
Read 2 more answers
Help Please..
statuscvo [17]

She should have to pay $380.25 in interest for the jet ski.

6 0
3 years ago
Given F(x) = V(3x – 12) + 2, what is the range in set notation?
iogann1982 [59]

Answer:

(-∞,∞)

Step-by-step explanation:

6 0
3 years ago
Unsure how to do this calculus, the book isn't explaining it well. Thanks
krok68 [10]

One way to capture the domain of integration is with the set

D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}

Then we can write the double integral as the iterated integral

\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx

Compute the integral with respect to y.

\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)

Compute the remaining integral.

\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}

We could also swap the order of integration variables by writing

D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}

and

\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy

and this would have led to the same result.

\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)

\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)

7 0
1 year ago
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