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3 years ago

15

(6/y+2) + (1/y-2) = 1

2 answers:

Ainat [17]3 years ago

5 0

Answer:

y = 7

Step-by-step explanation:

Solve for y:

7/y = 1

Take the reciprocal of both sides:

y/7 = 1

Multiply both sides by 7:

Answer: y = 7

Greeley [361]3 years ago

4 0

The answer is y=7 :))))))

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Kelsie sold digital cameras on her website. She bought the cameras for 465 each and included a 60% markup to get the selling pri

melamori03 [73]

Each camera costs $651 dollars, assuming I understood the question format correctly.
Find 60% of 465, which is 186. Because it is markup, add 186 to the original price. 465+186 = 651.
Therefore, the price is $651.

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4 years ago

Write the standard equation for the circle whose center is at (6,2) and whose radius is 2.

Grace [21]

Answer:

$(x - 6)^2 + (y - 2)^2 = 2^2$ .

Step-by-step explanation:

In general, let $(a,\, b)$ denote the center of a given circle. Let $r$ denote the radius of that circle. The standard equation for that circle would be:

$(x - a)^2 + (y - b)^2 = r^2$ .

The circle in this question is centered at $(6,\, 2)$ .

$6$ is the $x$ -coordinate of the center of this circle. $a = 6$ .
$2$ is the $y$ -coordinate of the center of this circle. $b = 2$ .

The radius of this circle is $2$ . Therefore, $r = 2$ .

Hence, the standard equation for this circle would be:

$(x - 6)^2 + (y - 2)^2 = 2^2$ .

8 0

3 years ago

Evaluate 10 to the negative 2nd power

storchak [24]

10 times 10 is equivalent to 100

7 0

3 years ago

Read 2 more answers

Which value, when multiplied by 5/16 results in a product of 1?

Sidana [21]

Multiply it by 16/5, or the reciprocal

6 0

3 years ago

I need help answering the question in the picture provided

adoni [48]

answer:

(a).

Equation of circle is <u>x²</u><u> </u><u>+</u><u> </u><u>y²</u><u> </u><u>-</u><u> </u><u>2</u><u>5</u><u> </u><u>=</u><u> </u><u>0</u>

(b).

(-5, 0) » yes

(√7, 1) » no

(-3, √21) » no

(0, 7) » no

Step-by-step explanation:

(a).

If centred at origin, centre is (0, 0)

General equation of circle:

${ \boxed{ \bf{ {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0}}}$

but g and f are 0:

${x}^{2} + {y}^{2} + c = 0 \\ but : \\ c = {g}^{2} + {f}^{2} - {r}^{2} \\ c = { - 25}$

8 0

3 years ago