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3 years ago

The maximum capacity of the school auditorium is 350 people. Enter an inequality

Mathematics

1 answer:

alexira [117]3 years ago

7 0

Answer:

Not enough info for an inequality

Step-by-step explanation:

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An isosceles triangle has a perimeter of 36 inches. Its base is 2

astra-53 [7]

The lengths of the sides of the triangle are 8, 8, 20

Explanation:

Given that the perimeter of an isosceles triangle is 36 inches.

The base of the triangle is $2\frac{1}{2}$ times longer than each of its legs.

We need to determine the lengths of the sides of the triangle.

<u>Lengths of the sides:</u>

Let x denote the lengths of the sides of the triangle.

The base of the triangle is given by

$2\frac{1}{2}x=\frac{5}{2}x=2.5x$

Perimeter of the isosceles triangle = Sum of the three sides of the triangle.

Thus, we have,

$36=x+x+2.5x$

$36=4.5x$

$8=x$

Thus, the length of the sides of the isosceles triangle is 8 inches.

Base of the triangle = $2.5(8)=20 \ inches$

Hence, the three sides of the isosceles triangle are 8, 8, 20

3 0

4 years ago

Solve the equation. Check your answer:<br><br> 14+6a-8=18

Ivanshal [37]

A=2...
Subtract 14 from both sides,then subtract 8 from both sides. Your left with
6a=12
6x1=6
6x2=12
A=2

5 0

3 years ago

graph PQR with vertices P(-2,3),Q (1,2), and R (3,-1) and its image after the translation ( x,y) -> ( x-2, y - 5)

Furkat [3]

Answer:

(x,y)(x-2,yt)

Step-by-step explanation:

(x,t)(x-2,yt)

we sniled it to the lete by 2 and then we snifted it down by

8 0

2 years ago

A bag contains blue marbles and red marbles, 56 in total. The number of blue marbles is 8 more than 2 times the number of red ma

Nikolay [14]

24 Blue marbles.

Explanation:

56 - 8 = 48

48 / 2 = 24

so if there are 24 blue marbles, 24 x 2 = 48, and then 8 more is 56.

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