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2 years ago

(x2 - x - 2) ÷ (x - 2)

Mathematics

2 answers:

Semmy [17]2 years ago

8 0

Answer:

x + 1

Step-by-step explanation:

(x² − x − 2) / (x − 2)

Factor the numerator. Using the AC method:

ac = (1)(-2) = -2

Factors of -2 that add up to -1 are -2 and 1

Divide by a and reduce: -2/1 and 1/1

The factors are (x − 2) and (x + 1)

(x − 2) (x + 1) / (x − 2)

Divide:

x + 1

kifflom [539]2 years ago

7 0

Answer:

(x+1)

Step-by-step explanation:

(x^2-x-2)/(x-2)=(x+1)

Because (x-2)(x+1)=x^2-2x+x-2=x^2-x-2.

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Find the value of x: A. X = 41.0 B. X = 122.0 C. X = 61.0 D. X = 120.5

ella [17]

Step-by-step explanation:

hope this will help you..

6 0

2 years ago

Continuation of the other six questions find the value of eggs and tangent

n200080 [17]

Answer:

Step-by-step explanation:

1) ?

I can't see what that one given angle is .. but it's that mystery angle subtracted from 180 give the angle BOA then you know that's an isosceles triangle again that the BOA is part of... so then just subtract BOA from 180 to find the two other angle of that triangle... they are the small so just divide your answer of 180-BOA /2 is the angle of each.. then since you know those to smaller angles subtract one from 90 to find the angle BAX

2) ( as we would read normally)

You're making this really tough on me.. I can just barely read the equations

I think it's 1+6x and 7x-3 . b/c they are the same length sides you can set those equal

1+6x = 7x -3

4+6x = 7x

4 = x

that worked out well :

for the tangent.. it's Tan(Ф)= Opp/ Adj

but I don't know which side they want to solve for.. I think you may have left off some of the instructions???? :/

ohh I think they really mean.. what's the length of the tangent lines .. that was confusing to me.. :/ just plug in 4 into x for either eq.

1+6(4) = 25

7(4)-3=25

tangent is 25

x-2 = 2x-10

x+8 = 2x

8 = x

again they made it work out easy :)

Then plug 8 into either equation to find the length of the tangent lines

8-2=10

tangent is 10

4) = 2) ??? they are the same question maybe you meant to put something else?

5 0

2 years ago

What equation represents a parabola with a focus of (0,4) and a directrix of y=2

scZoUnD [109]

We know that

Any point (x,y) on the parabola is equidistant from the focus and the directrix

Therefore,

focus (0,4) and directrix of y=2

√[(x−0)²+(y−4)²]=y−(2)

√[x²+(y-4)²]=y-2

x²+(y-4)²=(y-2)²

x²+y²-8y+16=y²-4y+4

x²=4y-12-----> 4y=x²+12----->y= (x²/4)+3

the answer is

y= (x²/4)+3

7 0

3 years ago

Lezlie hiked 3/8 of a mile on Monday. On Wednesday she hiked 3/6 of a mile. She hiked 3/4 of a mile on Friday. Use benchmark num

Gnoma [55]

She walked the most on Friday because you have to find the LCD. The LCD is 24.
For example, to change the denominator of 4 to 24, you must divide 24 divided by 4 which equals 6. Since your answer was 6, you have to multiply both numbers, which are 3 and 4, by 6.
So, 4 times 6 is 24. This is the denominator and 4 times 3 is 18. This is the numerator.
If you do the same with all of the fractions then you will have the fractions of, 18 over 24, 12 over 24, and 9 over 24.
Look at all the numerators (the number on top of the fraction) and decide which one is the biggest.
18 is the biggest and since the fraction 18 over 24 was originally 3 over 4, this means that she walked the most on Friday.
Hoped this helps!!

3 0

3 years ago

Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

7 0

2 years ago