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2 years ago

Expand and simplify: 5(x-3)+2(2x+1)

2 answers:

5 0

First you have to distribute.
5(x - 3) + 2(2x + 1)
5x - 15 + 4x + 2
Then you combine like terms.
9x - 13
This is your final answer. I hope this helps love! :)

adoni [48]2 years ago

5 0

5(x-3)+2(2x+1)
5x-15 +2(2x+1)
5x-15 +4x +2
9x-13

9x-13 is your answer

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Find an area of a triangle with a base of 25 feet and height of 8 feet. What’s the formula for a triangle?

marissa [1.9K]

Answer:

100 feet squared: can u plz give me the brainliest answer!!!!!!!!! plz!!!!!!!!!!!

Step-by-step explanation:

Area of triangle: base x height divided by 2.

25 x 8 = 200

200/2 = 100

Have a nice day!!!!!!!!!! :-)

4 0

2 years ago

Find two numbers whose sum is 44 and whose difference is 6.<br> smaller number<br> larger number

Anna [14]

X+ y = 44
x - y = 6
—————
2x = 50

x = 25
y = 44-x = 19

3 0

2 years ago

Do these two in the figure appear to be similar?

iVinArrow [24]

Hello,

Please, see the attached file.

Thanks.

4 0

3 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

3 0

1 year ago

Question 19 (5 points) ✔ Saved

zheka24 [161]

The order of temperature values become -16.7°C > -11.15°C > -7°C > 3°C And -16.7°C is the coolest temperature and 3°C is the warmest as compare to others.

According to the statement

we have given that the some values of the temperatures and we have to write it in the order from the coolest to the warmest means in increasing order.

So, For this purpose,

The given values of temperatures are:

3°C , -7°C , 2°F, 262 K

Firstly convert into one form

So, convert all values in Celsius form,

3°C = 3°C

-7°C = -7°C

2°F = -16.7°C

262 K = -11.15°C

Now arrange into the order from coolest to warmest

So, it will become

-16.7°C > -11.15°C > -7°C > 3°C

So, -16.7°C is the coolest temperature and 3°C is the warmest as compare to others.

The order of temperature values become -16.7°C > -11.15°C > -7°C > 3°C And -16.7°C is the coolest temperature and 3°C is the warmest as compare to others.

Learn about the increasing order here

brainly.com/question/14662181

#SPJ1

7 0

4 months ago