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3 years ago

This is for a Math test plz help me

Mathematics

2 answers:

Mrac [35]3 years ago

7 0

Answer:

Step-by-step explanation:

garik1379 [7]3 years ago

5 0

Answer:

Step-by-step explanation:

1*7=7+23=30

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Find and explain the error in the division problem below.

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If your aloud to change number it would be 95 Divided by 4 equals 23 because 95 divided by 4 makes 23

If not then 95 divided by 4 equals 21

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Simplify<br> 2a - 5 + 3a + 4

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Answer:

5a - 1

Step-by-step explanation:

add common terms

2a - 5 + 3a + 4

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X-1/3=6 what is this as an improper fraction

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Answer:6 1/3 or 19/3

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Write an equation to find the number of ounces n in any number of pounds p

nekit [7.7K]

Answer:

B. $n = 16p$

Step-by-step explanation:

number of ounces = n

number of pounds = p

There are 16 ounces in 1 pounds (unite rate: 1 pounds = 16 ounces)

Therefore, there would be n ounces in p pounds

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$\frac{16}{1} = \frac{n}{p}$

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8 0

2 years ago

Consider the function . For which intervals is f(x) positive? Check all that apply. (–∞, –3) (–3, –2) (–2, 2) (2, 3) (∞, 3)

Alborosie

<em>Note: As you missed to add the function, so after a little research I was able to find the function. It will anyways help you in terms of clearing your concept about the context.</em>

Answer:

For the intervals (–3, –2) and (2, 3), the function $\:f\left(x\right)=\frac{9-x^2}{x^2-4}$ will be positive. It is also clear from the graph attached below.

Step-by-step explanation:

Considering the function

$\:f\left(x\right)=\frac{9-x^2}{x^2-4}$

The intervals (∞, 3) and (–2, 2) will make the function negative. As 1 is between the interval (∞, 3) as well as between the interval (–2, 2).

For example, putting x = 1 in the function

$\:f\left(x\right)=\frac{9-x^2}{x^2-4}$

$f(x)=\frac{9-1^2}{1^2-4}$

$\mathrm{Apply\:rule}\:1^a=1$

$1^2=1$

$f(x)=\frac{9-1}{1-4}$

$\mathrm{Subtract\:the\:numbers:}\:9-1=8$

$f(x)=\frac{8}{1-4}$

$\mathrm{Subtract\:the\:numbers:}\:1-4=-3$

$f(x)=\frac{8}{-3}$

$f(x)=-\frac{8}{3}$

Thus, the values between the intervals (∞, 3) and (–2, 2) will make the function negative. It is also clear from the graph attached below.

Similarly, the values between the interval (–∞, –3) will also make $\:f\left(x\right)=\frac{9-x^2}{x^2-4}$ negative.

For example, putting x = -5 in the function

$\:f\left(x\right)=\frac{9-x^2}{x^2-4}$

$f(x)=\frac{9-\left(-5\right)^2}{\left(-5\right)^2-4}$

$f(x)=\frac{-16}{\left(-5\right)^2-4}$

$f(x)=\frac{-16}{21}$

$f(x)=-\frac{16}{21}$

Thus, the values between the interval (–∞, –3) will also make $\:f\left(x\right)=\frac{9-x^2}{x^2-4}$ negative. It is also clear from the graph attached below.

The values between the the interval (–3, –2) will make the function positive. For example, putting x = -2.5 in the function,

$\:f\left(x\right)=\frac{9-x^2}{x^2-4}$

$f(x)=\frac{9-\left(-2.5\right)^2}{\left(-2.5\right)^2-4}$

$f(x)=\frac{2.75}{\left(-2.5\right)^2-4}$

$f(x)=\frac{2.75}{2.25}$

$f(x)=1.22$

Thus, the values between the the interval (–3, –2) will make the function positive. It is also clear from the graph attached below.

Similarly, the values between the interval (2, 3) will also make $\:f\left(x\right)=\frac{9-x^2}{x^2-4}$ positive. For example, putting x = 2.5 in the function,

$\:f\left(x\right)=\frac{9-x^2}{x^2-4}$

$f(x)=\frac{9-\left(2.5\right)^2}{\left(2.5\right)^2-4}$

$f(x)=\frac{2.75}{2.5^2-4}$

$f(x)=\frac{2.75}{2.25}$

$f(x)=1.22$

Thus, the values between the the interval (2, 3) will make the function positive. It is also clear from the graph attached below.

Therefore, we can CONCLUDE that for the intervals (–3, –2) and (2, 3), the function $\:f\left(x\right)=\frac{9-x^2}{x^2-4}$ will be positive. It is also clear from the graph attached below.

Keywords: equation , interval, graph

Learn more about equation, graph and interval from brainly.com/question/12768877

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6 0

3 years ago

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