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2 years ago

How do I simplify 8(12+a) please quick

Mathematics

2 answers:

Kitty [74]2 years ago

5 0

8x12=96
8xa=8a
96+8a
you distribute

velikii [3]2 years ago

3 0

8(12+a)

expand

8*12+8*a=

96+8a

<u><em>Steps shown above</em></u>

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NikAS [45]

Answer:

2x-7

-14

x = 1

Step-by-step explanation:

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3 years ago

PLSSS HELPPP A football quarterback goes for a two-point conversion when the ball is within 10 yards of the end zone. During the

Evgen [1.6K]

Answer:

The probability of missing both two-point conversion attempts is 7.5%

Step-by-step explanation:

We are informed that the probability of missing the first attempt is 50% of the time. Furthermore, the probability of missing on the second attempt given that he missed the first attempt is 15% of the time

Now,the probability of missing on both the two-point conversion attempts will simply be given by the product of these two probabilities since the events are independent;

50%*15% = 0.5 * 0.15 = 7.5%

Therefore, the probability of missing both two-point conversion attempts is 7.5%

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2 years ago

Can you guys please help me

m_a_m_a [10]

The answer is B. First turn all the results from the spelling competition into fractions (or ratios).
Megan: 45/50 = 9/10
Justin: 27/30 = 9/10
Fernando: 84/120 = 7/10
Then, use this information to find which answer is correct.
Have a nice day! :)

4 0

3 years ago

A group consists of fivefive men and sevenseven women. FourFour people are selected to attend a conference.

exis [7]

Answer:

Step-by-step explanation:

4 people can be selected in ways=12 P4=12×11×10×9=11880

4 women can be selected in ways=7P4=7×6×5×4=840

$P=\frac{840}{11880} =\frac{7}{99}$

6 0

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

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3 years ago