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Nonamiya [84]
2 years ago
13

Complete the remainder of the

Mathematics
1 answer:
allsm [11]2 years ago
4 0

The corresponding values needed will be -23, -8, 7 and 22

<h3>Functions and Tables</h3>

Given the function y = 5x - 8

<h3>Get the domains and range</h3>

We are to get the range of the function for the given domains

If the value of x is -3

y = 5(-3) - 8

y = -15 - 8

y = -23

If the value of x is 0

y = 5(0) - 8

y = 0 - 8

y = -8

If the value of x is 3

y = 5(3) - 8

y = 15 - 8

y = 7

If the value of x is 6

y = 5(6) - 8

y = 30 - 8

y = 22

Hence the corresponding values needed will be -23, -8, 7 and 22

Learn more on tables of function here: brainly.com/question/3632175

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Mr Stevenson has 160 cows goats horses and chickens on his farm 19 percent of the animals are cows 28 percent goats 13 percent h
castortr0y [4]

Answer:

64 Chickens

Step-by-step explanation:

1. Add up the percents

19% + 28% + 13% = 60%

2. Subtract from 100%

100% - 60% = 40%

3. Multiply the total number of animals by the percent of chickens

160 * .4 = 64 Chickens



7 0
2 years ago
Read 2 more answers
How do u solve radius
kotykmax [81]

Answer:

To calculate the radius of a circle by using the circumference, take the circumference of the circle and divide it by 2 times π.

The two formulas that are useful for finding the radius of a circle are C=2*pi*r and A=pi*r^2.

The diameter is 2 times larger than the radius

Step-by-step explanation:

Hope this helped

:)

8 0
2 years ago
Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

Where,

A_0 = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = \frac{A_0}{2},

i.e.

\frac{A_0}{2}=A_0 e^{19k}

\frac{1}{2} = e^{19k}

0.5 = e^{19k}

Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

\implies k = \frac{\ln(0.5)}{19}\approx -0.03648

Now, if the substance to decay to 78​% of its original​ amount,

Then A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0

0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0
2 years ago
Find the prime factorization for the number 48 expressed in exponential notation.<br>Thank you.
STatiana [176]
48
12 * 4
6 * 2 * 2 * 2
3 * 2 * 2 * 2 * 2
it should be 3 * 2^4...but that is not an answer choice

3 * 2^2 * 4 is ur answer....apparently they didn't factor the 4


6 0
3 years ago
Again helppp! asap!
igomit [66]

Answer:

C

Step-by-step explanation:

because 30÷6 is 5 and that is what should be placed

3 0
3 years ago
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