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2 years ago

Match the graph with the correct equation. Please help!

Mathematics

1 answer:

galina1969 [7]2 years ago

6 0

Answer:

y = -x -2

Step-by-step explanation:

The graph has a slope of -1 and a y-intercept of -2.

The equation is therefore

y = -x -2

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Uestion 6 one out of every five people who buy ice cream buys vanilla ice cream. if a store sells 125 ice cream cones in one day

Snowcat [4.5K]

1/5 = .20

.2 x 125 = 25

25 Ice Cream cones would be Vanilla

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3 years ago

Given the point (18, -10), apply the rule and tell the image after the translation as an ordered pair with no spaces.

Phantasy [73]

Answer:

The answer is (7,-19)

Step-by-step explanation:

Remember, 18 is x, and -10 is y. The rule says to subtract 11 from 18, which would be 7. For y, you add -9 to -10, which would be -19.

Therefore, your answer should be (7,-19)

Hope this helps!

3 0

3 years ago

What is the equivilent of +367 on a numberline?

Mama L [17]

Answer:

367

Step-by-step explanation:

367 because it is positive

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2 years ago

The diagram below is divided into equal parts. Which ratio correctly compares the number of shaded sections to the total number

krek1111 [17]

Answer: 1/2 is correct.

Step-by-step explanation:

There are 8 total parts, and 4 of them are shaded. 4 is half of 8, therefore, your answer is 1/2.

Hope this helps!

8 0

3 years ago

Read 2 more answers

5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l

sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

$\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})$

Then we can rearrange and integrate

$\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50} \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}$

Then we have the model of A(t) like

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt reaches 50 lb is

$A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5$

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

7 0

3 years ago