3 years ago

Find the midpoint of the segment connecting (4,7) to (-2,3).

Mathematics

1 answer:

KonstantinChe [14]3 years ago

8 0

If you put 2 points on a graph paper and connect them, you can find the midpoint pretty easily

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What is the expression (a^2+b^2)^2 is equivalent to

netineya [11]

<span>(a^2+b^2)^2
since (a+b)^2=a^2+2ab+b^2, substitute a with a^2 and b with b^2, which yields:
a^4+2(a^2*b^2)+b^4</span>

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3 years ago

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Richard’s annual college expenses are expected to total $17,745. He will receive $5,320 in grants. How much will Richard need to

NISA [10]

The answer to your question is (c) $12,425, or the third option.

subtract how much he gets in grants from how much he owes

-Hope this helped!

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3 years ago

Solve for x. 4/x−3+2/x^2−9=1/x+3

Levart [38]

Answer:

$x=-\frac{17}{3}$

Step-by-step explanation:

we have

$\frac{4}{x-3}+\frac{2}{x^{2}-9}=\frac{1}{x+3}$

Solve for x

Applying difference of squares in the denominator of the second term in the left side

$\frac{4}{x-3}+\frac{2}{(x+3)(x-3)}=\frac{1}{x+3}$

Multiply both sides by (x+3)(x-3)

$4(x+3)+2=(x-3)$

Apply the distributive property in the left side

$4x+12+2=x-3$

Combine like terms left side

$4x+14=x-3$

Group terms

$4x-x=-3-14$

$3x=-17$

Divide by 3 both sides

$x=-\frac{17}{3}$

6 0

4 years ago

Use fraction strips to find each sum or difference simply if possible 6-22/3

Sergeeva-Olga [200]

The answer to ur question should be 133

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4 years ago

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This exercise involves the formula for the area of a circular sector. find the area of a sector with central angle 3π/5 rad in a

netineya [11]

The formula of interest is
A = (1/2)r²·θ
You have r = 9 m, θ = 3π/5. Then the area is
A = (1/2)(9 m)²(3π/5) = 24.3π m²

The sector area is ...
A ≈ 76.3 m²

8 0

3 years ago