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3 years ago

6

A gasoline service station offers 0.20 cents off the regular price per gallon every Tuesday. What equation relates the discounte

d price, d to the regular price, r on that day?

1 answer:

gavmur [86]3 years ago

5 0

D - 0.20 cents = r + 0.20 cents = d

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If F(x) = 5x - 7 and G(X) = x2 - 41, what is G(F(x))?

Viktor [21]

C is the answer to this

4 0

3 years ago

If a=2b^3 and b= -1/2c ^-2,express a in terms of c.

ElenaW [278]

Answer:

Part 1) $a=-\frac{1}{4c^6}$

Part 2) $a=-\frac{1}{4c^{-6}}$

Step-by-step explanation:

Part 1) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2}c^{-2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{1}{2}c^{-2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{1}{2}c^{-2})^{3}=2(-\frac{1}{2})^3(c^{-2})^{3}=2(-\frac{1}{8})(c^{-6})=-\frac{1}{4c^6}$

therefore

$a=-\frac{1}{4c^6}$

Part 2) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2c^{-2}}=-\frac{c^{2}}{2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{c^{2}}{2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{c^{2}}{2})^{3}=2(-\frac{c^{6}}{8})$

simplify

$a=-\frac{c^{6}}{4}$

therefore

$a=-\frac{1}{4c^{-6}}$

6 0

3 years ago

Solve 15x+3y=9,10x+7y=-4

TiliK225 [7]

Answer:

x = 1, y = -2

Step-by-step explanation:

15x + 3y = 9 (1)

10x + 7y = -4 (2)

3 × (2) - 2 × (1)

3 × (2)

30x + 21y = -12 (3)

2 × (1)

30x + 6y = 18 (4)

(3) - (4)

15y = -30

y = -30 / 15

y = -2

sub y = -2 into (1)

15x + 3y = 9

15x + (3 × -2) = 9

15x - 6 = 9

15x = 9 + 6

15x = 15

x = 15 / 15

x = 1

4 0

3 years ago

Can someone plz explain I would appreciate it

mafiozo [28]

Answer:

Angle of depression

Step-by-step explanation:

4 0

3 years ago

15. Before oil is added to a certain road surface material, the other ingredients are mixed. The mixture contains

mojhsa [17]

Answer:

sand = 3450 pounds

3/4 inch stones =6750 pounds

1 inch stones =4800 pounds.

Step-by-step explanation:

The mixture composition is :

45% three-quarter inch stones

32% inch stones and

sand for balance

The percent here is by mass of material used.

Weight of mixture is 15,000 pounds before oil addition.

Use the percent by weight for the two types of stones and calculate real values in pounds

Formula to apply is:

Mass percent = (mass of a material/ total mass of mixture) * 100

For 3/4 inch stones it will be:

$0.45= \frac{x}{15000} *100\\\\0.45=\frac{x}{150} \\0.45*150=x\\x=6750$

The 3/4 inch stones weigh 6750 pounds

The 1 inch stones will be

$32/100 * 15000 =4800$

The 1 inch stones weigh = 4800 pounds

Total for the two items : 4800+6750 =11550 pounds

The sand used weighs : 15000 - 11550 = 3450 pounds

8 0

3 years ago