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3 years ago

How do you solve (-1) ^2

Mathematics

2 answers:

likoan [24]3 years ago

4 0

(-1)^2

-negative and -negative= + positive

-1*-1 - two times ( because of the exponent)

Answer: 1

BabaBlast [244]3 years ago

3 0

Answer:

Step-by-step explanation:

You multiply -1 x -1 because it's squared and the negatives cancel out and 1 times 1 is 1

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3( x + 4y + 6) - 2(x + 8y + 3).

ser-zykov [4K]

Answer:

x-4y+12

Step-by-step explanation:

4 0

3 years ago

if f(1) = -5 and f(n) = f(n-1) +7, find the first four terms and the common difference of the sequence

Alex_Xolod [135]

Answer:

- 5, 2, 9, 16 and d = + 7

Step-by-step explanation:

to obtain the first four terms substitute n = 2, 3, 4 into the recursive formula

f(1) = - 5 ← given

f(2) = f(1) + 7 = - 5 + 7 = 2

f(3) = f(2) + 7 = 2 + 7 = 9

f(4) = f(3) + 7 = 9 + 7 = 16

common difference d = 16 - 9 = 9 - 2 = 2 - (- 5) = 7

6 0

3 years ago

Find 2× - 2y + 5z - 2x - y + 3z

natka813 [3]

<span>-3y+8z your welcome:)</span>

4 0

3 years ago

Can someone please help me factor and write the equations for these two problems ?

prisoha [69]

Answer:

Step-by-step explanation:

Asymptotes 3

g(x) = $\frac{3x}{x^2-3x-10}$

Factors of denominator will be,

x² - 3x - 10 = x² - 5x + 2x - 10

= x(x - 5) + 2(x - 5)

= (x + 2)(x - 5)

Therefore, factored form of g(x) will be,

g(x) = $\frac{3x}{(x + 2)(x - 5)}$

Asymptotes 4

h(x) = $\frac{(x-5)}{x^{2} + 14x + 40}$

= $\frac{x-5}{x^{2}+10x+4x+40}$

= $\frac{x-5}{x(x+10)+4(x+10)}$

= $\frac{x-5}{(x+10)(x+4)}$

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3 years ago

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Taya2010 [7]

Answer:

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Step-by-step explanation:

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3 years ago