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3 years ago

5

Which of the following transforms y = x' to the graph of y = (x + 5)22

1 answer:

Anni [7]3 years ago

6 0

Answer: A translation 5 units to the left... brainliest?

Step-by-step explanation: edge 2020

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A particular math test has eight questions and a perfect paper is worth 100 percentage points. What is the rate of change of the

Ostrovityanka [42]

100/8 = 12.5
The rate of change is 12.5

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2 years ago

Each one of the rectangles that make up the three dimensional figure is called a ___.

vovangra [49]

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Cuboid

Step-by-step explanation:

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3 years ago

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The probability of being struck by a unicycle is 3/50,000. What is the probability of not being struck?

ludmilkaskok [199]

49,997/50,000
Is the probability of not being struck by a unicycle

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3 years ago

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Grapes are 2.55 per pound and on two separate occasions Dana went to the store to

fomenos

Answer: $11.48

Step-by-step explanation:

First of all, you have to multiply 3.25 by 2.55 because the grapes are 2.55 per pound. That makes 8.2875 for the first occasion. Next, you have to multiply 1.25 by 2.55 for the second occasion. That makes 3.1875. You then have to add 8.2875 and 3.1875. You get 11.47500, and that rounded to the nearest cent is $11.48.

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2 years ago

X²+y²-6x+14y-1=0<br> and please show your work so I can learn

Eva8 [605]

Hello there,

I hope you and your family are staying safe and healthy during this winter season.

$x^2 + y^2 -6x+14y-1=0$

We need to use the Quadratic Formula*

$x =\frac{-b+\sqrt{b^2}-4ac }{2a}$ , $\frac{-b-\sqrt{b^2} -4ac }{2a}$

Thus, given the problem:

$a = 1, b=-6, c=y^2+14y-1$

So now we just need to plug them in the Quadratic Formula*

$x=\frac{6+2\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$ , $x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$

As you can see, it is a mess right now. Therefore, we need to simplify it

$x=\frac{6+2\sqrt{10-y^2-14y} }{2}$ , $x = \frac{6-2\sqrt{10-y^2-14y} }{2}$

Now that's get us to the final solution:

$x=3+\sqrt{10-y^2-14y}$ , $x=3-\sqrt{10-y^2-14y}$

It is my pleasure to help students like you! If you have additional questions, please let me know.

Take care!

~Garebear

3 0

2 years ago