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3 years ago

X(b+c)+3b+3c write the expression as a product of polynomials:

Mathematics

2 answers:

Luden [163]3 years ago

8 0

Answer:

(b+c)(X+3)

Step-by-step explanation:

X(b+c)+3b+3c

X(b+c)+3(b+c)

(b+c)(X+3)

kolbaska11 [484]3 years ago

5 0

Answer:

look at the picture above shown

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I'm stuck what is 15% of 32

Len [333]

In math, 'of' means 'multiplied by,' so 15% of 32 translates to '15% * 32.'
Now solve the equation:
x = 15% * 32
x = .15 * 32
x = 4.8

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3 years ago

The heights of adult females are normally distributed. If you were to construct a histogram of 40 randomly selected women, what

Anna11 [10]

Answer:

d. The histogram would be approximately bell-shaped, and the normal quantile plot would have data points have follow a straight-line pattern.

Step-by-step explanation:

Since the variable is normally distributed, the histogram of women's height should be approximately bell shaped (if the data was obtained form a random sample).

Again, the variable is normally distributed, therefore, the quantile plot should follow a straight line pattern (a diagonal line to be more precise).

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3 years ago

Carson drives to school the same way each day, and there are two independent traffic lights on his trip to school. he knows that

vodka [1.7K]

0.3x0.8=0.24 so 24% chance he wont have to stop i think

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3 years ago

Read 2 more answers

If a1 = 9 and and an-1 - 5 then find the value of a6.

SCORPION-xisa [38]

Answer:

a1 = 6 and an = an-1 - 5

a2 = a2-1 - 5 = a1 - 5 = 1

a3 = a3-1 - 5 = a2 - 5 = -4, etc

Step-by-step explanation:

8 0

3 years ago

Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 < 2pi. Show work please!

yan [13]

Answer:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}$

Step-by-step explanation:

You need 2 things in order to solve this equation: a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

$cos(2\theta)=1-2sin^2(\theta)$

so we will make that replacement, getting everything in terms of sin:

$sin(\theta)+1=1-2sin^2(\theta)$

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

$2sin^2(\theta)+sin(\theta)=0$

We can factor out the sin(theta), since it's common in both terms:

$sin(\theta)(2sin(\theta)+1)=0$

Because of the Zero Product Property, either

$sin(\theta)=0$ or

$2sin(\theta)+1=0$

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi. They are:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}$

The next equation needs to first be solved for sin(theta):

$2sin(\theta)+1=0$ so

$2sin(\theta)=-1$ and

$sin(\theta)=-\frac{1}{2}$

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval. They are:

$\theta=\frac{2\pi}{3},\frac{4\pi}{3}$

7 0

3 years ago