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Yakvenalex [24]
3 years ago
6

Dany has a business selling bracelets for 7.50 each. She only has enough material to make 5 bracelets. If x represents the numbe

r of bracelets sold and y represents the total cost, what is the range for this situation.
Mathematics
1 answer:
ziro4ka [17]3 years ago
6 0

Answer:

The range is [0,37.5]

Step-by-step explanation:

We start by writing the equation that links x and y

From the question;

y = 7.5x

Now, she has enough material to make 5 only

When we talk of range, we are referring to the values on the y-axis

So the lowest y value is 0

and since the highest number of bracelets is 5;

The total cost of the 5 is 7.5(5) = 37.5

So the range is [0,37.5]

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What is the prime factorization for 147
NemiM [27]
Prime Factorization = Divide by factors that are prime number

147 ÷  3 = 49
49 ÷ 7 = 7
7 ÷ 7 = 1
 
147 = 3 x 7 x 7

147 = 3 x 7²

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Answer: 147 = 3 x 7²
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Are these lines parallel, perpendicular, or neither? 3y=x+4 and 3x+y=1
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Answer:

perpendicular

Step-by-step explanation:

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3x + y = 1 in slope intercept form y = -3x +1

the slopes are negative reciprocals of each other so they are perpendicular

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Y=9r, r=r

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Helppppp my brain just doesn’t work today. I’ll brainlist.
s344n2d4d5 [400]

Answer:

Number 6 is 1/6

Number 7 is 5/6

Number 8 is 3/6 or 1/2

Step-by-step explanation:

6. \frac{1}{6} chance of landing on a 2 because there are 6 sides and the probability is equal.

There is a 1/6 chance it will land on 1, 2, 3, 4, 5, and 6. So 1/6 is the answer.

7. \frac{5}{6} chance of not landing on 5. There are 5 "non-5" numbers. So 5/6 is the answer.

8. Let's see the odd numbers: 1, 2, 3, 4, 5, 6.

1, 3, and 5 are odd. So that's half of the numbers so the answer would be \frac{3}{6} or simplified to \frac{1}{2}.

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3 years ago
please show on graph (with x and y coordinates) state where the function x^4-36x^2 is non-negative, increasing, concave up​
babunello [35]

Answer:

y'' =12x^2 -72=0

And solving we got:

x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}

We can find the sings of the second derivate on the following intervals:

(-\infty Concave up

x=-\sqrt{6}, y =-180 inflection point

(-\sqrt{6} Concave down

x=\sqrt{6}, y=-180 inflection point

(\sqrt{6} Concave up

Step-by-step explanation:

For this case we have the following function:

y= x^4 -36x^2

We can find the first derivate and we got:

y' = 4x^3 -72x

In order to find the concavity we can find the second derivate and we got:

y'' = 12x^2 -72

We can set up this derivate equal to 0 and we got:

y'' =12x^2 -72=0

And solving we got:

x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}

We can find the sings of the second derivate on the following intervals:

(-\infty Concave up

x=-\sqrt{6}, y =-180 inflection point

(-\sqrt{6} Concave down

x=\sqrt{6}, y=-180 inflection point

(\sqrt{6} Concave up

8 0
3 years ago
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