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Find two positive angles and two negative angles that are coterminal with the given angle and less than 1080 but greater than 72

ivann1987 [24]

Answer:

The correct answer is 450°, 810°, -270° and -630°.

Step-by-step explanation:

According to the given scenario, the calculation of the two positive angles and two negative angles i.e. coterminal is as follows:

The coterminal angles are the angles in which the difference could be 360 degrees or the multiples of 360 degrees

For 90 degrees, the two positive angles are

a. 90° + 360°

= 450°

450° + 360°

= 810°

b. The two negative angles are

= 90° - 360°

= -270°

-270° - 360°

= -630°

3 0

2 years ago

What is the value of the right rectangular prism

storchak [24]

Hello!

To find the area of a rectangular prism you do 2(lw + lh + wh)

Put in the values

2(9 * 2 + 9 * 6 + 2 * 6)

2(18 + 54 + 12)

2(84)

2 * 84 = 168

The answer is 168 cubic inches

Hope this helps!

6 0

2 years ago

Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

ANSWER:

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

SOLUTION:

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

8 0

2 years ago

Hey You YES YOU CAN ANSWER THIS PLEASE solve for T: 4 j= -9T

OLEGan [10]

-9T means -9 x T
So if T is equivalent to 4,
you should do:
-9 x 4 = -36
J= -36

8 0

3 years ago

An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 180180 engin

tiny-mole [99]

Answer:

$z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98$

$p_v =P(Z$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=4.6$ represent the sample mean

$\sigma=0.9$ represent the population standard deviation

$n=180$ sample size

$\mu_o =4.8$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if thetrue mean is below the specifications, the system of hypothesis would be:

Null hypothesis: $\mu \geq 4.8$

Alternative hypothesis: $\mu < 4.8$

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98$

P-value

Since is a one sided test the p value would be:

$p_v =P(Z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.

4 0

3 years ago