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2 years ago

8

ZB is 3x-5 BA is x+3

1 answer:

Akimi4 [234]2 years ago

3 0

If B is the midpoint of ZA, then ZB = <span>BA
</span>3x-5 = <span>x+3
3x - x = 3 + 5
2x = 8
x = 8/2
x = 4

</span>ZA = ZB + BA
= 3x-5 + x+3
= 3*4 - 5 + 4 + 3
= 12 - 5 + 7
= 14

Answer: <span>D. 14</span>

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2(a³+b²+11)+1(3+a³+b+11)

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Answer:

$2(a^3+b^2+11)+1(3+a^3+b+11)=3a^3+2b^2+b+36$

Step-by-step explanation:

I assume that you need simplification of the given expression.

The given expression is:

$2(a^3+b^2+11)+1(3+a^3+b+11)$

Using distributive property and multiplying 2 inside the parenthesis and 1 inside the other parenthesis. This gives,

$2(a^3+b^2+11)=2\times a^3+2\times b^2+2\times 11\\2(a^3+b^2+11)=2a^3+2b^2+22\\\\1(3+a^3+b+11)=1\times 3+1\times a^3+1\times b+1\times 11\\1(3+a^3+b+11)=3+a^3+b+11=a^3+b+14$

Therefore, $2(a^3+b^2+11)+1(3+a^3+b+11)$ is equal to:

$2a^3+2b^2+22+a^3+b+14$

Now, combining like terms using the commutative property of addition, we get:

$=(2a^3+a^3)+2b^2+b+(22+14)\\=3a^3+2b^2+b+36$

Therefore, the simplified form is $3a^3+2b^2+b+36$

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Step-by-step explanation:

we have

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Step-by-step explanation:

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<em>Horizontal asymptotes are found by equating numerator to 0 and solving for x</em>

<em>Vertical asymptotes are found by equating denominator to 0 and solving for x</em>

<em />

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<u>Vertical Asymptote:</u>

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