All categories

1 year ago

Find the area of the triangle below.

Mathematics

2 answers:

xenn [34]1 year ago

7 0

Answer:

Well t find the area of a triangle, you need to multiple Base x height. That gives you 43.56. Now you divide by 2. So the end product will give you 21.78. Those same rules apply to every time you are finding the area of a triangle.

Step-by-step explanation:

baherus [9]1 year ago

7 0

Answer: A

Step-by-step explanation: 7.8 x 4.4= 34.32 x .5= 17.16 = A

You might be interested in

Given two central angles in any two circles, the ratio of the arc length to the radius of each corresponding circle is equal if

krok68 [10]

First, we have
s1/r1 = s2/r2

The question also states the fact that
s/2πr = θ/360°

Rearranging the second equation, we have
s/r = 2πθ/360°

Then we substitute it to the first equation
s1/r1 = 2πθ1/360°
s2/r2 = 2πθ2/360°

which is now
2πθ1/360° = 2πθ2/360°

By equating both sides, 2π and 360° will be cancelled, therefore leaving
θ1 = θ2

6 0

2 years ago

Find the volume of each sphere. Round to the nearest tenth.

g100num [7]

Answer:

d. 463246.7

Step-by-step explanation:

V=4 /3πr^3=4/3·π·48^3≈463246.7

5 0

2 years ago

Ms.Liang is building a deck that is 2 2/9 yards long and 3 2/5 yards wide what is the area of her deck.

choli [55]

Answer:

length = 20/9 yards

breadth = 17/5 yards

area = length x breadth

= 20/9 x 17/5 = 7 5/9 sq. yards

6 0

2 years ago

Calculus 3 help please.

Reptile [31]

I assume each path $C$ is oriented positively/counterclockwise.

(a) Parameterize $C$ by

$\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}$

with $-\frac\pi2\le t\le\frac\pi2$ . Then the line element is

$ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt$

and the integral reduces to

$\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt$

The integrand is symmetric about $t=0$ , so

$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

(b) Parameterize $C$ by

$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{3^2+4^2} \, dt = 5\,dt$

and

$\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt$

Integrate by parts with

$u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}$

$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt$

and

$\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}$

8 0

8 months ago

In isosceles triangle ABC, AB=AC.If B=55, calculate A

den301095 [7]

Answer: ∠A = 70°

because ABC is an isosceles triangle and AB = AC

=> ∠B = ∠C = 55°

=> ∠A = 180-(∠B + ∠C) = 180° - 55°.2 = 70°

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers