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3 years ago

What is the equation of the line that is parallel to y−5=−13(x+2) and passes through the point (6,−1)?

Mathematics

1 answer:

BartSMP [9]3 years ago

7 0

Answer:

y=-13x + 78 I'm not sure what's up with those answer choices though

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4 times x with an exponent of 2 plus 2 times y with an exponent of 4. X=-3 y=2

Digiron [165]

Answer:

Step-by-step explanation:

Your equation is: $4*x^{2} +2 * y^{4}$

You also know that x = -3 and y = 2. You can rewrite the equation by substituting x and y with their actual values:

$4*(-3)^{2} +2 * 2^{4}$

First, you want to do the exponentiation (meaning you want to calculate $(-3)^{2}$ and $2^{4}$ parts). $(-3)^{2}$ equals 9 and $2^{4}$ equals 16.

So the equation now is:

$4*9+2*16$

Now you want to do multiplication. $4*9$ equals 36 and $2*16$ equals 32. So you're left with:

$36+32$

Which equals 68.

8 0

2 years ago

Write an expression to find the greatest difference:

Bumek [7]

Answer:

Try Khan Academy. They give lots of expression to find the greatest difference. It always helps me in this situation.

Step-by-step explanation:

4 0

2 years ago

How many grams of a 15 percent alcohol solution should be mixed with 40 grams of a 30 percent solution to obtain a

Akimi4 [234]

The number of grams is 20 grams

Let x represent how many grams of a 15 percent alcohol solution

15%x+30%.40/x+40

=25%

0.15x+0.4.30=0.25 (x+40)

0.15x+12=0.25x+10

0.15x-0.25x=10+12

-0.1x=-2

x=-2/-0.1

x=20

Inconclusion The number of grams is 20 grams

Learn more here:

brainly.com/question/24750329

4 0

2 years ago

a 23 gram sample of a substance thats a by-product of fireworks has a K-value of .13. whats the substances half life in days

nata0808 [166]

Given:
Initial mass, m₀ = 23 g
Decay constant, k = 0.13

Exponential decay obeys the equation
$m(t)=m_{0} e^{-kt}$
where t = time, days

For half life, the mass is m = m₀/2.
Therefore
$e^{-0.13t} = \frac{1}{2} \\ -0.13t = ln(0.5)\\t= \frac{ln(0.5)}{-0.13} =5.33$

Answer: The half life is 5.33 days

3 0

2 years ago

Read 2 more answers

A recent study found that the average length of caterpillars was 2.8 centimeters with a

pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

$\mu = 2.8, \sigma = 0.7$ .

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:

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

$Z = \frac{4 - 2.8}{0.7}$

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

4 0

1 year ago