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3 years ago

Please help its a test. explain and select all statements that are true

Mathematics

2 answers:

lapo4ka [179]3 years ago

7 0

B &D are Correct

B explanation: The variable is the 3, as the 3 is the amount per ride but we don’t know how many rides so you would multiply the unknown (x) by the 3, so this answer would be correct.

D explanation: if you replace the four with the x for both they both equal $28.

Otrada [13]3 years ago

5 0

Answer:

i think b and d

Step-by-step explanation: the $4.75 per mile and $3 per mile represent the "m" in y=mx+b. you have to multiply 4.75 by the number of miles travelled, and add that to the $9 additional fee to find the total amount charged. for the "driveyou" equation, you multiply $3 by the number of miles travelled, which is "x" in the equation. hope that helps, i'm new to brainly :)

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stealth61 [152]

16 hours has passed

6 0

3 years ago

3. The total cost for 9 bracelets, including shipping was $72. The shipping charge was $9. Define your variable and write an equ

ikadub [295]

Answer:

The equation that models the cost of each bracelet is $72=9x+9$ . Cost of each bracelet is $7.

Step-by-step explanation:

Let the cost of each bracelet is defined by the variable x.

Cost of 9 bracelet is 9x. The shipping cost is $9. Therefore the total cost of 9 bracelets, including shipping is

$\text{Total Cost}=9x+9$

The total cost for 9 bracelets, including shipping is $72.

$72=9x+9$

Subtract 9 from both sides

$72-9=9x+9-9$

$63=9x$

Divide both sides by 9.

$7=x$

Therefore the cost of each bracelet without shipping changes is $7.

8 0

3 years ago

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

$g(y) = \frac{y-5}{y^2-3y+15}$

According to the quotient rule:

$d(\frac{f(y)}{h(y)}) = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}$

Applying the quotient rule:

$g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}$

The values for which g'(y) are zero are the critical points:

$g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10$