3 years ago

Find the midpoint of AB given A (-1,4) and B (7,6)

Mathematics

1 answer:

lutik1710 [3]3 years ago

8 0

Seriously: add the x's divide by 2, add the y's divide by 2, bob's your uncle. you have the midpoint.

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you have a gift card worth $50 you want to buy several paperback books that cost $6 each write and solve an inequality to find t

VARVARA [1.3K]

$6x$5=$30

$50-$30=$20

So 5 books?

Or
50=6x5+x
x=20

3 0

3 years ago

According to the fundamental theorem of algebra, how many zeros does the function f(x)=4x^3-x^2-2x+1 have?

Ksivusya [100]

Since the maximum exponent of x is 3, there should be three roots.
Unless you have a root squared and multiplyed by a different root, you will only have two roots since two are the same.
Also, if all three roots are the same, you only have one root.

4 0

3 years ago

Please help me with this question.

Vesna [10]

Answer:

3 to 9

second is 2.5 to 15

Step-by-step explanation:

3 0

3 years ago

Hello, precalc, need help on finding csc

Ainat [17]

Recall the double angle identity for cosine:

$\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - 2 \sin^2(x)$

It follows that

$\sin^2(x) = \dfrac{1 - \cos(2x)}2 \implies \sin(x) = \pm \sqrt{\dfrac{1-\cos(2x)}2} \implies \csc(x) = \pm \sqrt{\dfrac2{1-\cos(2x)}}$

Since 0° < 22° < 90°, we know that sin(22°) must be positive, so csc(22°) is also positive. Let x = 22°; then the closest answer would be C,

$\csc(22^\circ) = \sqrt{\dfrac2{1-\cos(44^\circ)}} = \sqrt{\dfrac2{1-\frac5{13}}} = \dfrac{\sqrt{13}}2$

but the problem is that none of these claims are true; cot(32°) ≠ 4/3, cos(44°) ≠ 5/13, and csc(22°) ≠ √13/2...

3 0

2 years ago

Solve the equation for the indicated variable.<br><br> x=2y-2 for y

stira [4]

Answer:

2y - 2 = x

2y = x + 2

y = (x + 2)/2

3 0

3 years ago