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Alona [7]
2 years ago
10

How the heck do I do this I’m having a crisis there is a test on Thursday

Mathematics
1 answer:
iren [92.7K]2 years ago
4 0

Answer:

Common factor from the two pairs

n2+2n-n-2 or

n(n+2)-1(n+2)

Rewrote in factored form

n(n+2)-1(n+2)or

(n-1)(n+2)

Solution

(n-1)(n+2)

You might be interested in
A. 55<br> b. 70<br> c. 110<br> d. 125
morpeh [17]

Answer:

b

Step-by-step explanation:

7 0
2 years ago
If f(1)=6f(1)=6 and f(n)=nf(n-1)-4f(n)=nf(n−1)−4 then find the value of f(3)f(3).
alekssr [168]

Answer:

49

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Does y = 3/4 x + 6 and y = -3/4 x + 5 have one solution
Mumz [18]

Answer:

yes

Step-by-step explanation:

Given the 2 equations

y = \frac{3}{4} x + 6 → (1)

y = - \frac{3}{4} x + 5 → (2)

Substitute y = \frac{3}{4} x + 6 into (2)

\frac{3}{4} x + 6 = - \frac{3}{4} x + 5 ( add \frac{3}{4} to both sides )

\frac{3}{2} x + 6 = 5 ( subtract 6 from both sides )

\frac{3}{2} x = - 1 ( divide both sides by \frac{3}{2} )

x = \frac{-1}{\frac{3}{2} } = - \frac{2}{3} ← one solution

7 0
3 years ago
Read 2 more answers
Please HELPPP!!!!!!!!!!!!!!!!
DochEvi [55]

Answer:

40,44,48,52

Step-by-step explanation:

These are multiples of 4 because 4 can be multiplied into all of them like

4x11=44

4x10=40

4x12=48

4x13=52

3 0
2 years ago
I need help finding surface area for this shape
Alekssandra [29.7K]
So we got a rectangle base, 2 rectangle sides and 2 triangle sidess

ok so 6cm=height total


ok

so the trianglular sides
the base is4 and height is 6
2 triangles so
2 times 1/2 times 4 time 6=24 for both triangles

now the rectangle parts

the bottom bit is 4 by 12 or 48

the back one is 7 by 12 or 84

the front one is unknown because we don't know the side length of the front side (if we use pythagoran theorem and stuff we get that the side is √(65-8√13)) or about6.01, so I would say about 6)
so 6 times 12=72

total is

triangles+rectangles=
24+48+84+72=
228

about 228 square cm (if that front side length is 6)
8 0
3 years ago
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