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3 years ago

Well this is very hard and I need help passing do anybody know the answer to this question

Mathematics

1 answer:

dimulka [17.4K]3 years ago

3 0

Answer:

-1.9

Step-by-step explanation:

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How do I find The diameter of a 1 inch circular object

JulsSmile [24]

If you know the radius of the circle do radius times 2.

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3 years ago

You cut out a piece of me, and now I bleed internally

Crank

Answer: huh

Step-by-step explanation:

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3 years ago

Roisin is a price analyst for a food distributor. One week she put together bids for 680 cases, 507 cases, 830 cases, 391 cases,

sineoko [7]

Answer: The average number of cases per bid is <u>599</u>.

Step-by-step explanation:

Given: One week she put together bids for 680 cases, 507 cases, 830 cases, 391 cases, and 587 cases.

Total cases = 680 + 507 + 830 + 391+587

= 2995

Number of bids = 5

Now, average number of cases per bid = $\dfrac{\text{Total cases }}{\text{Number of bids}}$

$=\dfrac{2995}{5}\\\\=599$

Hence, the average number of cases per bid is <u>599</u>.

6 0

3 years ago

You have 19 digs for your current volleyball season. There are 5 games left in the season. You want to break your previous recor

viktelen [127]

19 + X > 26

X > 7

To break your record you need more than 7 digs

7 0

3 years ago

Find the value of the variable y, where the sum of the fraction 2/y-3 and 6/y+3 is equal to the quotient.

NISA [10]

Answer:

Here we need to solve:

$\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} }$

The sum of the fractions is equal to the quotient between the fractions.

Notice that the two values:

y = 3

y = -3

make the denominator equal to zero, so those values are restricted.

We can simplify the right side to get:

$\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} } = \frac{2*(y + 3)}{6*(y - 3)} = 3*\frac{y + 3}{y - 3}$

Now we can multiply both sides by (y - 3)

$(y - 3)*(\frac{2}{y - 3} + \frac{6}{y + 3 }) = 3*(y + 3)\\2 + 6*\frac{y -3}{y + 3} = 3*(y + 3)$

Now we can multiply both sides by (y + 3)

$(2 + 6*\frac{y -3}{y + 3})*(y + 3) = 3*(y + 3)*(y + 3)$

$2*(y + 3) + 6*(y - 3) = 3*(y + 3)*(y + 3)\\\\2*y + 6 + 6*y - 18 = 3*(y^2 + 2*y*3 + 9)\\\\8*y - 12 = 3*y^2 + 6*y + 33\\\\0 = 3*y^2 + 6*y + 33 - 8*y + 12\\\\0 = 3*y^2 - 2*y + 45$

First, let's see the determinant of that quadratic equation:

$D = (-2)^2 - 4*3*45 = -536$

We can see that it is negative, thus, there are no real solutions of the equation.

Thus, there is no value of y such that the origina equation is true,

6 0

3 years ago