All categories

2 years ago

I need help please tell the right answer

Mathematics

1 answer:

Roman55 [17]2 years ago

4 0

Answer: 47.6 is your answer hope this helped

plz make brainly

Step-by-step explanation:

You might be interested in

Can any one help me on this? Please thank you

nignag [31]

The area of a triangle is base length x height then divide all by 2,
so 12 x 10 /2
=60 ft²

5 0

3 years ago

The value of a house is increasing by $2500 per year. If it is worth $190,000 today, what will it be worth in four years?

sineoko [7]

Answer:

5000

Step-by-step explanation:

i dont know this answer but i try that

3 0

3 years ago

In ΔQRS, the measure of ∠S=90°, the measure of ∠Q=29°, and SQ = 5 feet. Find the length of QR to the nearest tenth of a foot.

goldfiish [28.3K]

Answer:

2.8 feet

Step-by-step explanation:

tan(29)=x/5

5tan(29)=x

2.8=x

8 0

3 years ago

Read 2 more answers

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

Which does NOT represent a function <br> a<br> b<br> c<br> d

mote1985 [20]

the answer is d
hope this helps

6 0

3 years ago