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3 years ago

Look at the image please helppp

Mathematics

1 answer:

balu736 [363]3 years ago

7 0

Answer:

Option A. one rectangle and two triangles

Option E. one triangle and one trapezoid

Step-by-step explanation:

step 1

we know that

The area of the polygon can be decomposed into one rectangle and two triangles

see the attached figure N 1

therefore

Te area of the composite figure is equal to the area of one rectangle plus the area of two triangles

$A=(8)(4)+2[\frac{1}{2}((8)(4)]=32+32=64\ yd^2$

step 2

we know that

The area of the polygon can be decomposed into one triangle and one trapezoid

see the attached figure N 2

therefore

Te area of the composite figure is equal to the area of one triangle plus the area of one trapezoid

$A=\frac{1}{2}(8)(4)+\frac{1}{2}((4+8)(8)=16+48=64\ yd^2$

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Please help me I'm trying to pull my grade up

Tcecarenko [31]

Answer:

Step-by-step explanation:

Because if you add 18x to both sides of the equation, you get 0 = -18

Step-by-step explanation:

When you combine like terms in the equation, you get -18x = -18x - 18. After adding 18x to both sides, you have 0 = -18. This is not true, which means that there are no solutions to the equation.

3 0

3 years ago

Read 2 more answers

Find all the solutions of the given equations in the interval [0,2pi) tan^3x=tanx and cos 3x=-cos3x

slamgirl [31]

$\tan^3x=\tan x$
$\tan^3x-\tan x=0$
$\tan x(\tan^2x-1)=0$
$\tan x(\tan x-1)(\tan x+1)=0$
$\begin{cases}\tan x=0\\\tan x-1=0\\\tan x+1=0\end{cases}$

$\tan x=\dfrac{\sin x}{\cos x}=0\implies \sin x=0\implies x=0,\pi$

$\tan x-1=0\iff\tan x=1\implies x=\dfrac\pi4,\dfrac{5\pi}4$

$\tan x+1=0\iff\tan x=-1\implies x=\dfrac{3\pi}4,\dfrac{7\pi}4$

- - -

$\cos3x=-\cos3x$
$2\cos3x=0$
$\cos3x=0\implies 3x=\dfrac\pi2,3x=\dfrac{3\pi}2\implies x=\dfrac\pi6,x=\dfrac\pi2$

This doesn't account for all the solutions, however; there are some values of $x$ that push $3x$ outside the interval $[0,2\pi)$ , so let's take a few more:

$3x=\dfrac{5\pi}2\implies x=\dfrac{5\pi}6$
$3x=\dfrac{7\pi}2\implies x=\dfrac{7\pi}6$
$3x=\dfrac{9\pi}2\implies x=\dfrac{9\pi}6$
$3x=\dfrac{11\pi}2\implies x=\dfrac{11\pi}6$

We can stop there, since the next candidate gives

$3x=\dfrac{13\pi}2\implies x=\dfrac{13\pi}6>2\pi$

7 0

3 years ago

The circumference of the bike tire above is 83.524 inches. What is the radius of the bike tire? (Use 3.14 for .) A. 262.27 in B.

Shtirlitz [24]

Answer:

Option <u>C. 13.3 in</u>.

Step-by-step explanation:

Here's the required formula to find the radius of the bike tire :

$\longrightarrow{\pmb{\sf{C_{(Circle)} = 2\pi r}}}$

C = circumference
π = 3.14
r = radius

Substituting all the given values in the formula to find the radius of the bike tire :

$\begin{gathered} \qquad{\longrightarrow{\sf{C_{(Circle)} = 2\pi r}}} \\ \\ \qquad{\longrightarrow{\sf{83.524 = 2 \times 3.14 \times r}}} \\ \\ \qquad{\longrightarrow{\sf{83.524 = 6.28 \times r}}} \\ \\ \qquad{\longrightarrow{\sf{83.524 = 6.28r}}} \\ \\ \qquad{\longrightarrow{\sf{r = \frac{83.524}{6.28}}}} \\ \\ \qquad{\longrightarrow{\sf{r = 13.3 \: in}}} \\ \\ \qquad \star{\underline{\boxed{\sf{\red{r = 13.3 \: in}}}}}\end{gathered}$