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3 years ago

8

What is the domain and range for f(x)=x^2

1 answer:

Lerok [7]3 years ago

7 0

Answer:

Domain: all real numbers

Range: y ≥ 0

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A standard pair of six-sided dice is rolled. What is the probability of rolling a sum less than 7? Express your answer as a frac

zzz [600]

Answer:

0.4167 is the probability of rolling a sum less than 7.

Step-by-step explanation:

We are given the following in the question:

Event: A standard pair of six-sided dice is rolled

A: rolling a sum less than 7

Sample space:

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

A:

{(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1)}

Formula:

$\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

$P(A) = \dfrac{15}{36} = 0.4167$

0.4167 is the probability of rolling a sum less than 7.

8 0

2 years ago

2 cars are 700 miles apart. They travel toward each other. One of the cars travels at 75mph, the other travels at 100 mph. How l

olga2289 [7]

Answer: C 4

Step-by-step explanation:

Use the formula: distance = rate x time

We can say that train 1 travels a distance of x, and train 2 travels a distance of 700 - x

The rate of train 1 is 75 mph, and the rate of train 2 is 100 mph

The time traveled for the two trains will be the same. We can represent that with the variable t.

We have the following equation for train 1:

x = 75t

For train 2, we have this equation:

700 - x = 100t

Use the Substitution Method by replacing x in the equation for train 2 with the value 75t.

700 - 75t = 100t

700 = 175t

700/175 = 4 hours.

It will take 4 hours for the two trains to meet.

6 0

2 years ago

Find x. Round your answer to the nearest tenth of a degree.

iren [92.7K]

Answer:

51.70

Step-by-step explanation:

3 0

2 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago

Solve the equation. If there is no solution, select no solution.

sergey [27]

Answer:

no solution to the question

6 0

3 years ago

Read 2 more answers