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2 years ago

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Billy starts counting backwards by 7, starting at 5907. when he reaches a single-digit number, he stops counting. what is the nu

mber he stops at ?
pls give me a correct method to do it. thx

best answer brainliest

1 answer:

Akimi4 [234]2 years ago

5 0

If you do 5907/7 you will get 843.85714
Do 843*7=5901.
Then do 5907-5901=6
The final answer you will get is 6.

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Add the two expressions.<br><br> 3z−4 and 2z + 5<br><br> Enter your answer in the box.

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Answer:

5z+1

Step-by-step explanation:

(3z-4)+(2z+5)

3z-4+2z+5

3z+2z-4+5

5z-4+5

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An investor has an account with stock from two different companies. Last year, his

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22.7%

Given that last year stock for company A was $7200
The stock for company B last year was worth $3510
Stock in company A decreased by 24%
This means the new value of stock for company A became;
(100-24)/100 *$7200
76/100*$7200
0.76*7200 =$5472
Stock in company B decreased by 20%
This means the new value of stock for company B became;
(100-20)/100 *$3510
80/100*$3510
0.8*$3510
$2808
Original investors stock value was = $7200+$3510 =$10710
New investors stock value is = $5472+$2808=$8280
Decrease in value of stock = $10710-$8280 =$2430
percentage decrease in stock value = decrease in stock/original value of stock *100%
=2430/10710 *100 =22.689
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Someone else had the same problem answered
So here it is

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2 years ago

____ [38]

Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 3x + x³

Taking differentiate

$\frac{d}{dx}\left(3x+x^3\right)$

$\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'$

$=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)$

solving

$\frac{d}{dx}\left(3x\right)$

$\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'$

$=3\frac{d}{dx}\left(x\right)$

$\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1$

$=3\cdot \:1$

$=3$

now solving

$\frac{d}{dx}\left(x^3\right)$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}$

$=3x^{3-1}$

$=3x^2$

Thus, the expression becomes

$\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)$

$=3+3x^2$

Thus,

f'(x) = 3 + 3x²

Given that f'(x) = 15

substituting the value f'(x) = 15 in f'(x) = 3 + 3x²

f'(x) = 3 + 3x²

15 = 3 + 3x²

switch sides

3 + 3x² = 15

3x² = 15-3

3x² = 12

Divide both sides by 3

x² = 4

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{4},\:x=-\sqrt{4}$

$x=2,\:x=-2$

Thus, the value of x will be:

$x=2,\:x=-2$

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That would be (6x-4)

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It should be noted that the hourly delivery rate is calculated by dividing the income made by the number of hours worked.

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Your information is incomplete. Therefore, an overview of the information will be given.

In order to get an hourly rate, it's important to divide the income that the person gets by the number of hours that the person worked.

In this case, let's say that the income is $20000 and the number of hours that was worked monthly was 240 hours, the hourly rate will be:

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