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2 years ago

Find the area of the shade region in the shown below.

Mathematics

1 answer:

Phoenix [80]2 years ago

6 0

Step-by-step explanation:

rectangle

= 15.3 × (8.2 - 5.7 )

= 15.3 × 2.5

= 38.25

triangle

= 1/2 × (15.3 - 5.1+6.7) × 5.7

= 1/2 × 3.5 × 5.7

= 1/2 × 19.95

= 9, 97

total = 38.25+9.97 = 48.22 ft

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Function 1: y = 4x + 5

vlabodo [156]

Y=mx+b
m=slope

given
y=4x+5
m=4=slope

2.
for the points (x1,y1) and (x2,y2), the slope of the line passing trhough those points is (y2-y1)/(x2-x1)
given
(1,6) and (3,10)
slope=(10-6)/(3-1)=4/2=2

function 1 slope is 4
function 2 slope is 2
4>2

the first one has greater rate of change

6 0

3 years ago

Read 2 more answers

During the first stages of an epidemic, the number of sick people increases exponentially with time. Suppose that at = 0 days th

nydimaria [60]

T = days passed

r = rate of growth

by 0 day, or t = 0, there are 2 folks sick,

$\bf \qquad \textit{Amount for Exponential Growth}
\\\\
A=P(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &2\\
P=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &0\\
\end{cases}
\\\\\\
2=P(1+r)^0\implies 2=P\cdot 1\implies 2=P\qquad \boxed{A=2(1+r)^t}$

by the third day, t = 3, there are 40 folks sick,

$\bf \qquad \textit{Amount for Exponential Growth}
\\\\
A=P(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &40\\
P=\textit{initial amount}\to &2\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &3\\
\end{cases}
\\\\\\
40=2(1+r)^3\implies 20=(1+r)^3\implies \sqrt[3]{20}=1+r
\\\\\\
\sqrt[3]{20}-1=r\implies 1.7\approx r\qquad \boxed{A=2(2.7)^t}$

how many folks are there sick by t = 6? $\bf \stackrel{that~many}{A=2(2.7)^6}$

8 0

3 years ago

Hi answer for this please

miv72 [106K]

Answer:

x+79degree + 35degree = 180 degree ( being the sum of all angle of the triangle )

or, x + 114 degree = 180 degree

or, x = 180degree - 114degree

or, x = 66degree

hence, x = 66 degree.

3 0

2 years ago

Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0

eduard

If you're using the app, try seeing this answer through your browser: brainly.com/question/2787701

_______________

$\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}$

Square both sides:

$\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}$

$\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\
\mathsf{17\,sin^2\,\theta=1}\\\\
\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}$

Since $\mathsf{sin\,\theta}$ is positive, you can discard the negative sign. So,

$\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}$

Substitute this value back into $\mathsf{(i)}$ to find $\mathsf{cos\,\theta:}$

$\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

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2 years ago

4(3x+2)=4(2x+3)+4x how many solution are their

baherus [9]

Answer:

Step-by-step explanation:

0=4 ther is just one solution

7 0

2 years ago

Read 2 more answers

Find the area of the shade region in the shown below.​

Find the area of the shade region in the shown below.