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2 years ago

What is the answer? -4 + x = -26

Mathematics

2 answers:

algol132 years ago

3 0

Answer:

x = -30

Step-by-step explanation:

Pavel [41]2 years ago

3 0

Answer: x = -22

Step-by-step explanation:

1) -4 + x = -26

2) Carry the -4 to the other side. It will turn the -4 into a +4.

You will end up with x = -26 + 4.

Remember that variables go to the left side of the "=" and numbers stay to the right.

3) x = -22

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Point R lies on the directed line segment from L (-8,-10) to M (4,-2) and partitions the segment in the ratio 3 to 5. What are t

Dvinal [7]

Answer:

R = (- 3.5, - 7 )

Step-by-step explanation:

Using the Section formula

$x_{R}$ = $\frac{3(4)+5(-8)}{3+5}$ = $\frac{12-40}{8}$ = $\frac{-28}{8}$ = - 3.5

$y_{R}$ = $\frac{3(-2)+5(-10)}{3+5}$ = $\frac{-6-50}{8}$ = $\frac{-56}{8}$ = - 7

Thus coordinates of R = (- 3.5, - 7 )

7 0

3 years ago

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A plane travels at a constant speed. It takes 6 hours to travel 3360 miles.

ivolga24 [154]

Answer:

A-560

B-5600

Step-by-step explanation:

to find the answer for A we need to divide 3360 by 6 which gives us 560.

check: 560 x 6 = 3360

To find the answer for B we need to multiply the answer for A (560) by 10 which gives us 5600

3 0

3 years ago

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Write the inequality: 3 less than n is no more than -8

rusak2 [61]

Answer:

n - 3 ≤ -8

Step-by-step explanation:

n - 3 ≤ -8

3 less than n: n - 3

is no more than: ≤ - 8

5 0

2 years ago

If a /b = p /q, then (a + b )/b = _______. (p + b )/q (p + p)/q (p + q)/q (p + a)/q

maxonik [38]

Hope this helps you!

5 0

3 years ago

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Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

7 0

2 years ago

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