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choli [55]
1 year ago
12

The population of a certain city can be modeled by P = 130,000(1.018)t.

Mathematics
1 answer:
stepan [7]1 year ago
3 0

Answer:

D.the population change is inconsistent

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Bruh- can someone pls help me out kindly I will give 13 points
ycow [4]

Answer:

\huge\boxed{\texttt{Yes. Explanation for why.}}

Step-by-step explanation:

If we have the equation y = \frac{1}{2}x - 4 and we want to test if the point (4,–2) satisfies the equation, we can plug the x and y values into the equation and see if they make the equation true.

4 is the x value, -2 is the y.

-2 = (\frac{1}{2}\cdot4)-4

-2 = 2 - 4

-2 = -2

So this does satisfy the equation.

Hope this helped!

7 0
3 years ago
Read 2 more answers
A cardboard box 6 units high, 4 units wide, and 3 units long can hold 80 pounds of brown rice. A second cardboard box has the sa
iren [92.7K]
I hope this helps but use lingith × with × higth so you would × 6 ×4 ×3
3 0
3 years ago
Greatest common factor 48x^5,39x^6
lora16 [44]
4,853,096 is the number
6 0
3 years ago
A quarter of the sum of a number and 20 is equivalent to 30. Find the number.
Alisiya [41]

Answer:

X = 100

Step-by-step explanation:

X + 20 over 4 = 30

1/4x + 5 = 30

-5               -5

1/4x = 25

multiply both sides by 4

x = 100

4 0
3 years ago
A phone company offers two monthly plans. Plan A costs $10 plus an additional $0.15 for each minute of calls. Plan B costs $30 p
Mkey [24]

First, you need to write to expressions to model each situation:

Plan A: 10+0.15x

Plan B: 30+0.1x


Next, set the expressions equal to each other and solve for x:

10+0.15x=30+0.1x

<em>*Subtract 0.1x from both sides to isolate the variable*</em>

10+0.05x=30

<em>*Subtract 10 from both sides*</em>

0.05x=20

<em>*Divide both sides by 0.05*</em>

x=400


The plans would have the same cost after 400 minutes of calls.


To find how much money the plans cost at 400 minutes, plug 400 into either expression.  We'll use Plan A:

10+0.15(400)

10+60

70


The plans will cost $70.


Hope this helps!

3 0
3 years ago
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