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klasskru [66]
3 years ago
9

Again and again help me on this ixl basically grinding points 101

Mathematics
1 answer:
saveliy_v [14]3 years ago
5 0
F is independent , s is dependent
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8y-8(y-1)=6-(16y-4)-10 Solve the linear equation ?
slamgirl [31]

Answer:

y = -1/2

Step-by-step explanation:

First distribute.

8y - 8y + 8 = 6 - 16y + 4 - 10

8 = -16y

Divide by -16.

y = -1/2

6 0
3 years ago
A saleswoman earns 60% commission on all that she sells. Last month she sold $800 worth of merchandise. How much commission (In
Softa [21]

Answer:

$480

Step-by-step explanation:

From the above question, we are told that:

A saleswoman earns 60% commission on all that she sells. Last month she sold $800 worth of merchandise.

The amount of commission (In dollars) she earned last month is calculated as:

60% of $800

= 60/100 × $800

= $480

Therefore, the amount of commission shoe earned last month in dollars = $480

3 0
3 years ago
Please help me Thanks <3
Gnom [1K]
The answer is number two since in y=Mx+c, C is the y intercept. This means that 52 is the y intercept. Therefore, Jordan is charged $52 for each gigabyte of data she uses.
8 0
3 years ago
Ratios ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ
FromTheMoon [43]

Answer:

A)9/8

B) 9/17

Step-by-step explanation:

There are 17 total and 8 new so subtract 8 from 17 and u get 9. 17 total. 8 new. 9 used.

3 0
3 years ago
Find all relative extrema of the function. Use the Second Derivative Test where applicable. (If an answer does not exist, enter
Lapatulllka [165]

Answer:

Relative minimum: \left(-\frac{5}{2}, -\frac{33}{4}\right), Relative maximum: DNE

Step-by-step explanation:

First, we obtain the First and Second Derivatives of the polynomic function:

First Derivative

f'(x) = 2\cdot x + 5 (1)

Second Derivative

f''(x) = 2 (2)

Now, we proceed with the First Derivative Test on (1):

2\cdot x + 5 = 0

x = -\frac{5}{2}

The critical point is -\frac{5}{2}.

As the second derivative is a constant function, we know that critical point leads to a minimum by Second Derivative Test, since f\left(-\frac{5}{2}\right) > 0.

Lastly, we find the remaining component associated with the critical point by direct evaluation of the function:

f\left(-\frac{5}{2} \right) = \left(-\frac{5}{2} \right)^{2} + 5\cdot \left(-\frac{5}{2} \right) - 2

f\left(-\frac{5}{2} \right) = -\frac{33}{4}

There are relative maxima.

6 0
3 years ago
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