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2 years ago

15

7th grade math help me plzzz

1 answer:

Nuetrik [128]2 years ago

5 0

Answer:

i think it is 94

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10a^2b^0 for a=–3, b=–8

NeTakaya

Answer:

<h2>90</h2>

Step-by-step explanation:

Substitute a = -3 and b = -8 to the expression 10a²b⁰:

10(-3)²(-8)⁰ = 10(9)(1) = 90

a⁰ = 1 for any real number

7 0

2 years ago

Read 2 more answers

I need major help pretty pls this so different

crimeas [40]

First put a bubble around 5x-7. Now you plug that into the problem. -3x-2
(5x-7)=12. now times the -2 by 5x which is -10x. and then times the -2 by -7 which is 14. so now you have -3x-10x+14=-12. add the -3x and -10x. -13x. Add the 14 over. -13x=2. Now dived by -13, which gives us x=-0.15

Now you take that a plug it into the -3x-2y=-12. So that it looks like this -3(-0.15)-2y=-12

Now times the -3 by -0.15 to give us 20 -2y=-12. Now 'kill' the 20, so that it looks like this -2y=-32. Now you divide by -2 and get y=16.

Now your ordered pair is (-0.15,16)

7 0

2 years ago

How do i write, Two increased by three equals the quotient of ten and two. Use x to represent any unknown number?

inn [45]

Answer:

$2+3=\frac{10}{2}$

Step-by-step explanation:

we know that

The phrase " Two increased by three equals the quotient of ten and two" is equal to the algebraic expression

$2+3=\frac{10}{2}$

$5=5$ -----> is true

7 0

2 years ago

Consider the two functions. F(x)=x^2-8x+7

Tpy6a [65]

(X-7)(x-1)

Minimal x values=7,1
Not the same x value
7 is the greater minimal value

7 0

2 years ago

Select all the expressions that are equivalent to (2)^n+³

eimsori [14]

Answer:

The expressions which equivalent to $(2)^{n+3}$ are:

$4(2)^{n+1}$ ⇒ B

$8(2)^{n}$ ⇒ C

Step-by-step explanation:

Let us revise some rules of exponent

$a^{m}$ × $a^{m}$ = $a^{m+n}$
$(a^{m})^{n}$ = $a^{m*n}$

Now let us find the equivalent expressions of $(2)^{n+3}$

A.

∵ 4 = 2 × 2

∴ 4 = $2^{2}$

∴ $(4)^{n+2}$ = $(2^{2})^{n+2}$

- By using the second rule above multiply 2 and (n + 2)

∵ 2(n + 2) = 2n + 4

∴ $(4)^{n+2}$ = $(2)^{2n+4}$

B.

∵ 4 = 2 × 2

∴ 4 = 2²

∴ $4(2)^{n+1}$ = 2² × $(2)^{n+1}$

- By using the first rule rule add the exponents of 2

∵ 2 + n + 1 = n + 3

∴ $4(2)^{n+1}$ = $(2)^{n+3}$

C.

∵ 8 = 2 × 2 × 2

∴ 8 = 2³

∴ $8(2)^{n}$ = 2³ × $(2)^{n}$

- By using the first rule rule add the exponents of 2

∵ 3 + n = n + 3

∴ $8(2)^{n}$ = $(2)^{n+3}$

D.

∵ 16 = 2 × 2 × 2 × 2

∴ 16 = $2^{4}$

∴ $16(2)^{n}$ = $2^{4}$ × $(2)^{n}$

- By using the first rule rule add the exponents of 2

∵ 4 + n = n + 4

∴ $16(2)^{n}$ = $(2)^{n+4}$

E.

$(2)^{2n+3}$ is in its simplest form

The expressions which equivalent to $(2)^{n+3}$ are:

$4(2)^{n+1}$ ⇒ B

$8(2)^{n}$ ⇒ C

3 0

3 years ago