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umka2103 [35]
3 years ago
10

How do i complete this with quadratics?

Mathematics
1 answer:
Bond [772]3 years ago
7 0
I hope this helps you



log6 [x. (2x-7)]=log6 (6^2)


x. (2x-7)=36



2x^2-14x-36=0

2x +4



x -9


(2x+4)(x-9)=0


x= -2


x=9





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brand X pizza is having a special coupon offer. if you buy two large pizzas, you get $2.00 off the total price before tax. a lar
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Answer:

Very handy question to know the answer to. I know of no place on the planet that does not have a tax of some sort.

The tax only applies when money changes hands. It does not apply to any discounts

So if you take the deal, the cost is

Cost of 2 pizzas = normal price - the discount.

Cost of 2 pizzas = 2*10.50 - 2.00

Cost of 2 pizzas = 21 - 2

Cost of 2 pizzas = 19.00 dollars.

Not you add on the sales tax. You can do this one of two ways, the long way or the short way.

Long way

Find the sales tax

Tax = % * cost

Tax = (7.5/100) * 19

Tax = 1.425 don't round yet.

Total Cost = purchase + tax

Total Cost = 19.00 + 1.425 = 20.425 Now round. This is a real problem. It depends on what you have been told to do about rounding when 5 is the last digit. I round up. So my answer would be 20.43 <<<<< Answer.

Short Way

Cost = (Price - discount)*(1 + %/100)

Cost = (21 - 2)*(1+ 7.5/100)

Cost = 19 * (1.075)

Cost = 20.425 = 20.43

Step-by-step explanation:

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What is 5 times 5 because i had got it wrong
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Answer:

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Step-by-step explanation:

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Evalaute the following without using a calculator: <br> cos(13pi/6)
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the\ cos\ function\ is\ periodic:\ \ \ the\ period=2 \pi\\\\cos(2 \pi + \alpha )=cos \alpha \\\\------------------------\\\\cos\bigg{(} \frac{\big{13 \pi }}{\big{6}}\bigg{)}=cos\bigg{(} 2 \pi +\frac{\big{ \pi }}{\big{6}}\bigg{)}=cos\bigg{(} \frac{\big{ \pi }}{\big{6}}\bigg{)}= \frac{\big{ \sqrt{3} }}{\big{2}}
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x+y  ≤ 20

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3 0
3 years ago
Can you help me find t value in this problem?
Eduardwww [97]

we have a maximum at t = 0, where the maximum is y = 30.

We have a minimum at t = -1 and t = 1, where the minimum is y = 20.

<h3>How to find the maximums and minimums?</h3>

These are given by the zeros of the first derivation.

In this case, the function is:

w(t) = 10t^4 - 20t^2 + 30.

The first derivation is:

w'(t) = 4*10t^3 - 2*20t

w'(t) = 40t^3 - 40t

The zeros are:

0 = 40t^3 - 40t

We can rewrite this as:

0 = t*(40t^2 - 40)

So one zero is at t = 0, the other two are given by:

0 = 40t^2 - 40

40/40 = t^2

±√1 = ±1 = t

So we have 3 roots:

t = -1, 0, 1

We can just evaluate the function in these 3 values to see which ones are maximums and minimums.

w(-1) = 10*(-1)^4 - 20*(-1)^2 + 30 = 10 - 20 + 30 = 20

w(0) = 10*0^4 - 20*0^2 + 30    = 30

w(1) =  10*(1)^4 - 20*(1)^2 + 30 =  20

So we have a maximum at x = 0, where the maximum is y = 30.

We have a minimum at x = -1 and x = 1, where the minimum is y = 20.

If you want to learn more about maximization, you can read:

brainly.com/question/19819849

6 0
2 years ago
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