All categories

3 years ago

In ΔHIJ, the measure of ∠J=90°, JI = 48, IH = 73, and HJ = 55. What is the value of the tangent of ∠H to the nearest hundredth

Mathematics

1 answer:

babunello [35]3 years ago

3 0

Answer:

The value of the tangent of ∠H to the nearest hundredth is 41.11⁰

Step-by-step explanation:

Given;

∠J=90°

JI = 48

IH = 73

HJ = 55

If we sketch this triangle, we will observe that ΔHIJ is a right angled triangle, with hypotenuse as "IH", opposite as "JI" and adjacent as "HJ".

To determine the value of the tangent of ∠H, we apply the following formula;

$Tan \ H = \frac{Opposite \ side}{Adjacent \ side} = \frac{JI}{HJ} \\\\Tan \ H =\frac{48}{55} \\\\Tan \ H = 0.8727\\\\H = Tan^{-1} (0.8727)\\\\H = 41.111^o\\\\H = 41.11^o (nearest \ hundredth)$

Therefore, the value of the tangent of ∠H to the nearest hundredth is 41.11⁰

You might be interested in

Jagroop is building a dock at his cottage. The length of the doc is 3 times the width, plus 2. Determine a simplified expression

MissTica

Answer:

Step-by-step explanation:

Let length = y width = x

y = 3x + 2

Perimeter = Sum of all sides (or sum of both lengths and both widths)

2y + 2x

2(3x + 2) + 2x

6x + 4 + 2x

8x + 4

7 0

2 years ago

Please help :) I’m really in need of assistance and would appreciate it.

BaLLatris [955]

Answer:

$D) 0.5x^3y + 4x^2 y^2 - 2xy^3$

Step-by-step explanation:

Set up an equation to model the situation,

$(-0.5xy)(x^2 + 8xy - 4y^2)$

Distribute, multiply each term in the parenthesis with the term outside of the parenthesis. Don't forget the laws of exponents, when one multiplies two exponents, with the same base, then one adds the two exponents. A number with no exponential term is two the first degree meaning that it is to the power of one.

$-0.5x^3 y - 4x^2 y^2 + 2xy^3$

4 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

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