I got HBIA buh ig I got it wrong

If PQ=8 and Q lies at -13 where could P be located?

choli [55]

Answer:

In the given figure the point on segment PQ is twice as from P as from Q is. What is the point? Ans is (2,1).

Step-by-step explanation:

There is really no need to use any quadratics or roots.

( Consider the same problem on the plain number line first. )

How do you find the number between 2 and 5 which is twice as far from 2 as from 5?

You take their difference, which is 3. Now splitting this distance by ratio 2:1 means the first distance is two thirds, the second is one third, so we get

4=2+23(5−2)

It works completely the same with geometric points (using vector operations), just linear interpolation: Call the result R, then

R=P+23(Q−P)

so in your case we get

R=(0,−1)+23(3,3)=(2,1)

Why does this work for 2D-distances as well, even if there seem to be roots involved? Because vector length behaves linearly after all! (meaning |t⋅a⃗ |=t|a⃗ | for any positive scalar t)

Edit: We'll try to divide a distance s into parts a and b such that a is twice as long as b. So it's a=2b and we get

s=a+b=2b+b=3b

⇔b=13s⇒a=23s

7 0

3 years ago

Please zoom in on the question <br> I will mark brainliast

erma4kov [3.2K]

B graph of function

4 0

3 years ago

Two containers designed to hold water are side by side, both in the shape of a cylinder. Container A has a diameter of 12 feet a

Brrunno [24]

Answer:

Step-by-step explanation:

The student needs to check all algebraic and mathmatical calculations for errors and typos. I'm old and have been prone to making mistakes.

Cylinder A DIAMETER 12 ft and Height 13 ft

Cylinder B DIAMETER 10 ft and Height 16 ft

After pumping How much water remains in cylinder A

Volume of a cylinder = π(radius)²h is the normal form since they provided the diameter I will use Volume of a cylinder = π(diameter/2)²h

Water remaining in A = Volume of A - Volume of B =

VolA - VolB = π(D for A/2)²H of A - π(D for B/2)²H of B

writing if a little more condensed

VolA - VolB = π(D for A/2)²H of A - π(D for B/2)²H of B

Va - Vb = π(Da/2)²Ha - π(Db/2)²Hb

= π [(Da²/2²)Ha - (Db²/2²)Hb] factored out the π

= π/4 [ Da²Ha - Db²Hb] factored out the (1/2)²

I can't think of any other algebraic steps

Va - Vb = Water Remaining in Container A = π/4 [ Da²Ha - Db²Hb]

= π/4 [ (12²)(13) - (10²)(16)]

= π/4 [ (144)(13) - (100)(16)]

= π/4 [ 1872 - 1600 ]

= π/4 [272]

= π [ 272/4 ]

= π [ 136 / 2]

= π [ 68 ]

= π (68)

= 213.6 ft³ rounded to the nearest tenth

I checking my answer by using V = πr²h r = d/2

Va - Vb = π [ra²ha - rb²hb]

= π [6²(13) - 5²(16)]

= π [(36)(13) - (25)(16)]

= π [ 468 - 400]

= 68 π

= 213.6 ft³ I got the same answer

= π

test

8 0

3 years ago

The sum of two consecutive odd integers is 316. Find the two odd integers.

Tcecarenko [31]

1, 3, 5, 7, 9 ......
notice, there are consecutive odd integers, every other number
skip one, another, skip one, another and so on

so, from 1 to 3, is 1 + 2 =3
5 is 3 +2
7 is 5+2
and so on

let us pick an odd integer, hmmm say "a"
to find the next odd one, it'd be "a + 2" of course :)

we know their sum is 316
so $\bf a+(a+2)=316\impliedby \textit{solve for "a"}$

once you found "a", the next one is, well, "a + 2" :)

5 0

3 years ago

Find the area of the shaded region of the trapezoid.

Talja [164]

The answer is B. have a good night

7 0

2 years ago