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3 years ago

PLEASE HELP WITH MATH! Quick problem. :)

Mathematics

2 answers:

Maru [420]3 years ago

8 0

6^2 = 36
5 x 6 = 30
36 - 30 = 6
6 - 3 = 3
The answer is 3

docker41 [41]3 years ago

7 0

The answer is three.

6^2 - 5(6) -3

36 - 30 - 3

6 - 3
3

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Aleonysh [2.5K]

$f(x) = (x - \dfrac{2}{9} )(x + \dfrac{1}{2} )$

$\text {When } f(x) = 0 :$

$(x - \dfrac{2}{9} )(x + \dfrac{1}{2} ) = 0$

$(x - \dfrac{2}{9} ) = 0 \text { or } (x + \dfrac{1}{2} ) = 0$

$x = \dfrac{2}{9} \text { or } x = -\dfrac{1}{2}$

3 0

3 years ago

Help please! A. B. C. D.

elixir [45]

Answer:

b = 175; which is the initial # of pages to start

Step-by-step explanation:

you can find the slope to find b or y-intercept (x=0) for equation

y=mx+b

to find m = (y2-y1)/(x2-x1)

(x1,y1) = (9, 100)

(x2, y2) = (15, 50)

m = (50-100) / (15-9) = -50/6 = -25/3

let use point (15, 50)

y=mx+b

50 = -25/3*x + b

50 = (-25/3)*(15) + b

50 = -125 + b

b = 175; which is the initial # of pages to start

3 0

3 years ago

The ratio of children to adults at family movie night is 5 to 4. A total of 72 people showed up at this weeks movie night. How m

N76 [4]

The answer is 40 I think

7 0

3 years ago

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

3 years ago

Read 2 more answers

Russell bought a $1,240 sound system on an installment plan. The installment plan included a 20% down payment and 18 monthly pay

KonstantinChe [14]

The down payment is $248
The total amount of monthly payments is 62

7 0

2 years ago