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Crank
2 years ago
8

Khan academy, i dont understand the concept

Mathematics
2 answers:
statuscvo [17]2 years ago
8 0

Khan academy, i dont understand the concept

g= 22

AURORKA [14]2 years ago
8 0

Answer:

g=22

Step-by-step explanation:

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Please need help i need help i need help
xeze [42]

Set each expression from the parentheses equal to 0 separately like so:

Equation 1: x - 4 = 0

Equation 2: -5x + 1 = 0

Now for each equation solve for x!

Equation 1:

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

x = 4

Equation 2:

-5x +( 1-1) = 0 - 1

-5x/ -5 = -1 / -5

x = 1/5

Check:

(4 - 4)(-5*4 + 1)

(0) (-20 + 1)

(0) (-19)

0 = 0 -------------------------> correct!

(1/5 - 4)(-5 * 1/5 + 1)

(-19/5)(-1 + 1)

(-19/5)(0)

0 = 0 -------------------------> correct!

smaller x = 1/5

larger x = 4

Hope this helped!

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3 years ago
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How many 2's must be multiplied for the product to be a number between 100 and 200
Advocard [28]
50. 2x50 = 100. To be safe I would go with any number Between 50 and 100
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kramer
#5 is 4, #6 is graph A, and #7 is graph B. I hope this helped, Please mark as brainliest because I am 3 more away from the next level.
7 0
3 years ago
A lawyer charges a $75 consultation fee and then $120 per hour thereafter.
Ainat [17]

Answer:

(a) The dependant veriable is the $120 per hour and the independant veriable is the $75 consultaion fee.
(b) The table of vaules for this would look like this,
\frac{C}{T} \frac{75}{0} \frac{195}{1} \frac{315}{2} \frac{435}{3} \frac{555}{4}, where C = cost and T = time.
(c) The relationship between C and T is linear, beacause with the more hours you take the cost increases.
(d) Yes, it is sensible because the points would be in line due to the steady increase of the hourly rate.

(e) For every hour spent the cost increases 120 dollars.
(f) The fixed cost is 75 dollars for consultation and the veriable cost is 120 per hour.

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2 years ago
Explain the answer and I'll add you as brand list
choli [55]

Answer:

how to find it out

Step-by-step explanation:

on the line that starts at x=-6 and ends at x=6.Find the spot where the y value is the highest

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