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3 years ago

(y-7)^2=2y^2-4y+73 solve for y

Mathematics

1 answer:

masya89 [10]3 years ago

7 0

Answer:

y=-4 or y=-6

Step-by-step explanation:

y^2-14y+49=2y^2-4y+73

y^2+10y+24=0

(y+4)(y+6)=0

y=-4 or y=-6

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11. The student council is comparing prices for their semi-formal dance. They compare banquet. Hall A charges $40 per person. Ha

Natalka [10]

Answer:

TC (A) = 40x , TC (B) = 500 + 20x

Step-by-step explanation:

Let the number of students be = x

Hall A Total Cost

Relationship Equation, where TC (A) = f (students) = f (x) 40 per person (student) = 40x

Hall B Total Cost

Relationship Equation, where TC (B) = f (students) = f (x) 500 fix fee & 20 per person (student) = 500 + 20x

4 0

3 years ago

A scientist removed a sample of 39.1 grams of a chemical from a container. The sample was 5 3/4 grams less than 3/10 of the tota

Oksanka [162]

The first thing we must do for this case is to define variables:

x: the total mass of the chemical in the container

y: a sample of a chemical from a container

We have the following equation:

y = (3/10) x - 5 3/4

Then, for y = 39.1 we have:

39.1 = (3/10) x - 5 3/4

Clearing x:

(3/10) x = 39.1 + 5 3/4

(3/10) x = 44.85

x = (10/3) * (44.85)

x = 149.5 grams

Answer:

the total mass in grams of the chemical in the container before the scientist removed the sample of 39.1 grams was:

x = 149.5 grams.

3 0

3 years ago

Simplify √9 x 3√27 please answer

irakobra [83]

Answer:

I got $27\sqrt{3}$ which is 46.76537.... in decimal form.

Step-by-step explanation:

4 0

3 years ago

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65

erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 2.65

$\sigma$ = standard deviation = 0.85

n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

P(X bar <= 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z < 2.06) = 0.98030

P(X bar <= 2.65) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3) = 0.98030 - 0.5 = 0.4803 .

8 0

4 years ago

30 POINTS FOR SIMPLE QUESTION

Komok [63]

-4 = y + 3

y = -4x - 3

Solve for Y in the first equation:

-4 = y +3

Subtract 3 from both sides:

y = -7

Now using the addition method:

rewrite the second equation as 4x+y = -3

Multiply y = -7 by - 1 to get -y = 7

Now add the two together:

4x +y = -3

-y =7

4x = 4

Divide each side by 4:

x = 4/4 = 1

Then replace x in one of the equations to solve for y which is -7

X = 1, Y = -7

5 0

3 years ago