All categories

3 years ago

Find the measure of the indicated angle to the nearest degree.

Mathematics

2 answers:

Yuliya22 [10]3 years ago

8 0

Answer:

Step-by-step explanation:

tan(theta)=perpendicular/base

tan(theta)=3/4

theta=arctan(3/4)=37

Dima020 [189]3 years ago

7 0

Answer:

37°

Step-by-step explanation:

for this question you have to use the tan ratio since the opposite of that angle has been given as 3 and the adjacent has also been given which is 4..I will represent the unknown angle using x so,

tan x=opposite/adjacent

tanx=3/4

tanx=0.75

x=tan inverse of 0.75

x=36.9 or 37°

I hope this helps

You might be interested in

How do I solve this equation using long division?

pychu [463]

2x-3

remainder 3

hope the picture explains it

3 0

3 years ago

Please help!!! Refer to the spinner. Find P (even or not shaded). A. 2/3 B. 1/2

scZoUnD [109]

It’s bbbbbbbbbbbbbbbbbbbb

5 0

3 years ago

Read 2 more answers

Find an equation of the tangent line to the curve 2(x^2+y^2)2=25(x^2−y^2) (a lemniscate) at the point (3,1)

Sholpan [36]

$\bf 2[x^2+y^2]^2=25(x^2-y^2)\qquad \qquad 
\begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 3}}\quad ,&{{ 1}})\quad 
\end{array}\\\\
-----------------------------\\\\
2\left[ x^4+2x^2y^2+y^4 \right]=25(x^2-y^2)\qquad thus
\\\\\\
2\left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[2x-2y\frac{dy}{dx} \right]
\\\\\\
2\left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=50\left[x-y\frac{dy}{dx} \right]
\\\\\\
$

$\bf \left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[x-y\frac{dy}{dx} \right]
\\\\\\
4x^3+4xy^2+4x^2y\frac{dy}{dx}+4y^3\frac{dy}{dx}+25y\frac{dy}{dx}=25x
\\\\\\
\cfrac{dy}{dx}[4x^2y+4y^3+25y]=25x-4x^3+4xy^2
\\\\\\
\cfrac{dy}{dx}=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}\impliedby m=slope$

notice... a derivative is just the function for the slope

now, you're given the point 3,1, namely x = 3 and y = 1

to find the "m" or slope, use that derivative, namely $f'(3,1)=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}$

that'd give you a value for the slope

to get the tangent line at that point, simply plug in the provided values
in the point-slope form

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad
\begin{cases}
x_1=3\\
y_1=1\\
m=slope
\end{cases}\\ \qquad \uparrow\\
\textit{point-slope form}$

and then you solve it for "y", I gather you don't have to, but that'd be the equation of the tangent line at 3,1

6 0

3 years ago

If n = 4, all of the following expressions are equivalent except _____.

Lena [83]

Answer:

Its 16

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Worth 98 points hElp

BartSMP [9]