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sesenic [268]
2 years ago
13

Factor y2 + 11y + 10.

Mathematics
1 answer:
strojnjashka [21]2 years ago
4 0

Answer:

(y+10)(y+1)

Step-by-step explanation:

y2+10y+y+10

y(y+10)+1(y+10)

(y+10)(y+1)

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Find the slope of a line perpendicular to 2x-y-16
Alekssandra [29.7K]

put the equation 2x - y = 16 in the form of y = mx + c

2x - y = 16

2x = 16 + y

y = 2x - 16

the slope of this line is 2. the slope of a line perpendicular to it would be the negative reciprocal of 2. in other words, it would multiply with 2 to give -1.

you can form this equation with that info

2x = -1

x = -1/2

OR

you can flip and change the sign (numerator) of 2/1

2/1

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6 0
3 years ago
What is 89 divide into 4356
ValentinkaMS [17]
Divide 89 by 4356
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6 0
3 years ago
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The first discount on a camera was 18%. The second discount was 20%. After these two discounts the price was $328. What was the
Artyom0805 [142]

Answer:

$500

Step-by-step explanation:

We can find the original price of the camera through a proportion. A proportion is an equation where two ratios or fractions are equal. The ratios or fractions compare like quantities.

<u>Second Discount</u>

20% off means we paid 80%. We know we paid $328 of some price.

\frac{80}{100} =\frac{328}{y}

I can now cross-multiply by multiplying numerator and denominator from each ratio.

80y=100(328)\\80y=32800

I now solve for y by dividing by 80.

\frac{80y}{80} =\frac{32800}{80} \\y= $410

The price after the first discount was $410.

<u>First Discount</u>

We will repeat the steps above with $410. 18% off means we paid 82%.

\frac{82}{100} =\frac{410}{y}

I can now cross-multiply by multiplying numerator and denominator from each ratio.

82y=100(410)\\82y=41000

I now solve for y by dividing by 82.

\frac{82y}{82} =\frac{41000}{82} \\y= $500

The original price was $500.


5 0
3 years ago
Find the value of x.
sveticcg [70]

If 98 - 32 = 66,

then x = 66°

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HELP! <br><br> Which relation is a function?<br><br> Select all that apply.
Shtirlitz [24]

Answer:

A,B,D

Step-by-step explanation:

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3 years ago
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