Write an equivalent expression for 88 + 16x + 8 if x = 4 is the solution to both expressions the same

Find the radius of the circle whose equation is x^2 + y^2 = 4.

ludmilkaskok [199]

Answer:

Radius = 2.

Step-by-step explanation:

We have been given equation of the circle which is $x^2 + y^2 = 4$ .

Now we need to find the radius of the given circle $x^2 + y^2 = 4$ .

To find that let's compare with the formula of the circle with standard equation of the circle $x^2 + y^2 = r^2$ , where r is the radius.

We get $r^2 = 4$

take square root of both sides.

$r=2$

Hence radius of the given circle is r = 2.

3 0

3 years ago

Which statement is true?

4vir4ik [10]

The correct answer is D because you cannot have a fraction as the base of a log

hope this helps :)

if you need further information just ask

4 0

4 years ago

I need some help please. Write a function rule to describe each statement...

hichkok12 [17]

You could see the pattern

it cost 3.8 cent to print 1 dollar,

It cost 2 x 3.8 cents to print 2 dollar

it cost 3 x 3.8 cents to print 3 dollar

First equation
f(x) = 3.8 x

Same with second equation,
f(x) = 15x

for equation 3, it would be :

y = 1.50 x - 0.8x

Hope this helps

4 0

3 years ago

Express the following expression in partial fractions 9x -7/2x-3x +1

oksian1 [2.3K]

Answer:

2 1/2x + 1

Step-by-step explanation:

9x -7/2x-3x +1 = 9x - 3 1/2x - 3x + 1 = 2 1/2x + 1

6 0

3 years ago

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 13.2 years, and standard deviation of

s344n2d4d5 [400]

Answer:

There is a 91.15% probability that their mean life will be longer than 12 years.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 13.2 years, and standard deviation of 3.1 years. This means that $\mu = 13.2, \sigma = 3.1$ .