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bagirrra123 [75]
3 years ago
10

Number 5 please help

Mathematics
1 answer:
Nadusha1986 [10]3 years ago
5 0

15k+1100=12k+20015k+1100=12k+200 \:
15k+1100−12k=200


3k+1100=200

3k=200-1100

3k= -900

k = -900/3

k = -300

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Katyanochek1 [597]
5 is the correct answer
4 0
3 years ago
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If Holly gets 14 % off a shirt , what percent of the original price will still pay​
scoundrel [369]

Answer:

Holly will still pay 86% of the original price.

Step-by-step explanation:

Given

  • Holly gets 14% off a shirt, so
  • Discount = 14%

To determine

What percent of the original price will still pay.

All we need is to subtract the percentage of discount from 100.

i.e.

100 - Discount = 100 - 14 = 86%

The amount left over would be what remains of the original price after Holly gets 14 % off a shirt.

Thus, Holly will still pay 86% of the original price.

3 0
3 years ago
12<br> 4<br> What is the value of x?<br> O 3/8<br> O 1<br> O<br> 8/3<br> C<br> 8
Tanya [424]

Answer:

the value of x is 48

Step-by-step explanation:

because if x=12 and y=4 than you have to do this:

-) (x+y) x/y

(12+4) times 12/4

4/6 times 2/41

than it equals 48

this took me so long ;-;

7 0
3 years ago
What are roots asymtotes and end behavior of (x-2)(x+3)
harina [27]
Roots:

x-2 = 0 

x = 2

x+3 = 0

x = -3

asymptotic:

none 

lim (x-2)(x+3) = (inf -2)(inf+3 ) = inf
x ->inf

lim (x-2)(x+3) = (-inf-2)(-inf+3) = inf
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8 0
4 years ago
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Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]
Zina [86]

Answer: Absolute minimum: f(-1) = -2\sqrt{6}

              Absolute maximum: f(\sqrt{12.5}) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = t\sqrt{25-t^{2}}

f'(t) = 1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)

f'(t) = \sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }

f'(t) = \frac{25-2t^{2}}{\sqrt{25-t^{2}} } = 0

For this function to be zero, only denominator must be zero:

25-2t^{2} = 0

t = ±\sqrt{2.5}

\sqrt{25-t^{2}} ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = -\sqrt{2.5} and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = t\sqrt{25-t^{2}}

f(-1) = -1\sqrt{25-(-1)^{2}}

f(-1) = -\sqrt{24}

f(-1) = -2\sqrt{6}

f(\sqrt{12.5}) = \sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}

f(\sqrt{12.5}) = 12.5

f(5) = 5\sqrt{25-5^{2}}

f(5) = 0

Therefore, absolute maximum is f(\sqrt{12.5}) = 12.5 and absolute minimum is

f(-1) = -2\sqrt{6}.

8 0
3 years ago
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