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skad [1K]
3 years ago
6

If a case of chips has 27 bags in it. How many cases would they need for 410 bags of chips?

Mathematics
1 answer:
kodGreya [7K]3 years ago
3 0

Answer:

they would need 15 cases for 410 bag of chips

Step-by-step explanation:

410/7= about 15

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Vertex: (3,-65) axis of Symmetry: x = 3

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Write an equation in​ slope-intercept form of the line that passes through the given point and is perpendicular to the graph of
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Answer:

y = -5x - 17

Step-by-step explanation:

Linear equations, or lines, that are perpendicular to each other have opposite reciprocals for the value of slope.  For the given line, y = 1/5x - 2, the slope is 1/5 and the opposite reciprocal is -5.

Given the value of the slope and a given point on the second line, you can solve for 'b':

y = mx + b

8 = (-5)(-5) + b

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-17 = b

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16. Solve the system of equations by graphing.<br> y = x + 8<br> y = - x- 6
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Answer:

Step-by-step explanation:

x + 8 = -x - 6

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(-7, 1)

5 0
3 years ago
40% is NOT equivalent to....? <br> A. 0.4 <br> B.2/5 <br> C.40/100 <br> D. 0.04 <br> E. None correct
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6 0
3 years ago
Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for
KIM [24]

Answer:

a) The recurrence formula is P_n = \frac{109}{100}P_{n-1}.

b) The general formula for the population of Tacoma is

P_n = \left(\frac{109}{100}\right)^nP_{0}.

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write P_0 = 200 000. For the population in 2001 we will use P_1, for the population in 2002 we will use P_2, and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that P_0 is equal to P_1 plus 9% of the population of 2000. Notice that this can be written as

P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0.

In 2002, we will have the population of 2001, P_1, plus the 9% of P_1. This is

P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1.

So, it is not difficult to notice that the general recurrence is

P_n = \frac{109}{100}P_{n-1}.

b) In the previous formula we only need to substitute the expression for P_{n-1}:

P_{n-1} = \frac{109}{100}P_{n-2}.

Then,

P_n = \left(\frac{109}{100}\right)^2P_{n-2}.

Repeating the procedure for P_{n-3} we get

P_n = \left(\frac{109}{100}\right)^3P_{n-3}.

But we can do the same operation n times, so

P_n = \left(\frac{109}{100}\right)^nP_{0}.

c) Recall the notation we have used:

P_{0} for 2000, P_{1} for 2001, P_{2} for 2002, and so on. Then, 2016 is P_{16}. So, in order to obtain the approximate population of Tacoma in 2016 is

P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062

d) In this case we want to know when P_n>400000, which is equivalent to

(1.09)^{n}P_0>400000.

Substituting the value of P_0, we get

(1.09)^{n}200000>400000.

Simplifying the expression:

(1.09)^{n}>2.

So, we need to find the value of n such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

n\ln(1.09)>\ln 2.

So, n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693.

So, the population of Tacoma should exceed the 400 000 by the year 2009.

8 0
3 years ago
Read 2 more answers
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