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3 years ago

Express the product of 2x^2+6x-8 and x+3 in standard form

1 answer:

7 0

The answer is 2x³ + 12x² + 10x - 24

(2x² + 6x - 8)(x + 3) = x(2x² + 6x - 8) + 3(2x² + 6x - 8) =
= 2x³ + 6x² - 8x + 6x² + 18x - 24 =
= 2x³ + (6x² + 6x²) + (-8x + 18x) - 24 =
= 2x³ + 12x² + 10x - 24

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Whats the answer two -9=7-x

garri49 [273]

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3 years ago

Read 2 more answers

At the neighborhood grocery, 5 pounds of salmon cost $49. How much would it cost

cluponka [151]

Answer:

$46.06

Step-by-step explanation:

Divide $49 by 5

9.8

So it costs $9.80 for 1 pound of salmon

Multiply $9.80 by 4.7

$46.06

Hope this helped :)

7 0

3 years ago

Graph f(x)=|x−2|+3 .<br><br> Use the ray tool to graph the function.

maks197457 [2]

Answer:

The graph will be 2 units away from the origin on positive $x-axis$ and three units upward from the origin towards $y-axis$ .

Step-by-step explanation:

Here is a graph attached with it.

To graph $\left | x-2 \right |+3$ we know that positive $\left | x \right |$ is a $V$ shaped from the origin.

Key points:

To move rightward there must be a negative inside the parentheses.
And to move upward we must have positive $b = constant$ .

If we have to move towards $x-axis$ then we must have negative inside it.

And if we have to move upward in $y-axis$ positive we must have positive constant value.

So the graph will be three units upward and two units rightward with a V-shaped ray.

4 0

3 years ago

The Rocky Mountain News (January 24, 1994) indicated that the 20-year mean snowfall in the Denver/Boulder region is 28.76 inches

ycow [4]

Answer:

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

Step-by-step explanation:

20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.

At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So

$H_0: \mu = 28.76$

At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

$H_1: \mu > 28.76$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

28.76 is tested at the null hypothesis:

This means that $\mu = 28.76$

Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.

This means that $\sigma = 7.5, X = 33, n = 32$ .

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}$

$z = 3.2$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level $\alpha = 0.05$ .