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3 years ago

Tell whether the following two triangles can be

Mathematics

1 answer:

zhuklara [117]3 years ago

3 0

Answer:

B. No, the two triangles don't have

corresponding sides marked congruent.

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Exam Guidelines

Ganezh [65]

Answer:

it's A because it's not B or C

7 0

1 year ago

X=(y)/(y-3) make y the subject

Liula [17]

Answer:

$\displaystyle y=\frac{3x}{x-1}$

Step-by-step explanation:

<u>Solving Equations</u>

We have the expression:

$\displaystyle x=\frac{y}{y-3}$

And it's required to make y the subject.

It can be done by isolating y on the left side of the equation and the rest of the expression on the right side.

The first step is removing the denominator by multiplying both sides by y-3:

$\displaystyle (y-3)x=(y-3)\frac{y}{y-3}$

Simplifying the right side:

$\displaystyle (y-3)x=y$

Remove the parentheses:

$yx-3x=y$

Move the y's to the left side and the rest of the expression to the right side:

$yx-y=3x$

Factor out y:

$y(x-1)=3x$

Divide by x-1:

$\mathbf{\displaystyle y=\frac{3x}{x-1}}$

7 0

2 years ago

What is 26.50 rounded to the nearest hundredth

Pavlova-9 [17]

Well,

26.50

In this case, the hundredths place is the 0.

Since there are no more digits after that, 26.50 must already be rounded to the nearest hundredth.

26.50 is the final answer.

6 0

2 years ago

Employees at a clothing store recieve 20% discounts on what they purchase at that store. How much will an employer save on a $45

mamaluj [8]

Easiest way to work it out is to divide by ten, that gives 10% of your original number. Multiply that by two to get 20% and minus that from the original number to get the discount.
450/10 = 45 which is 10% of 450.
20% of 450= 10% x 2 or 45 x 2
therefor 20% of 450 is 90.
450-90 = the discount the shoppers recieve.

4 0

3 years ago

arrange the expressions in the correct sequence to rationalize the denominator of the expression -(2)/(\sqrt(x+y-2)-\sqrt(x+y+2)

cupoosta [38]

We have to rationalize the denominator:
$\frac{-2}{ \sqrt{x+y-2} - \sqrt{x+y+2} } = \\ \frac{-2}{ \sqrt{x+y-2} - \sqrt{x+y+2} }* \frac{ \sqrt{x+y-2}+ \sqrt{x+y+2} }{ \sqrt{x+y-2}+ \sqrt{x+y+2} }= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2}) }{x+y-2-(x+y+2)}= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2}) }{x+y-2-x-y-2}= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2} }{-4}= \\ \frac{ \sqrt{x+y-2}+ \sqrt{x+y+2} }{2}$

6 0

2 years ago

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