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2 years ago

Can someone please help me in finding the area of this shape? (Also, this is the whole question, I didn't cut anything out)

Mathematics

2 answers:

Valentin [98]2 years ago

6 0

<span>I guess the answer is 96m square because you need to multiply length by width</span>

victus00 [196]2 years ago

5 0

To find the area, you have to find the area of 2 different shapes... because you have a rectangle that has a semicircle cut out of it, you have to first find the area of the whole rectangle, then subtract the area of the semicircle. The area of the whole rectangle would be...

A= length x width
A = 12m x 8m = 96 m^2

Now we need to find the area of the semicircle. C is the midpoint of the semicircle and since the circle fits perfectly in the width of the rectangle, we know it's diameter is 8m.

The area of a circle is found with the equation A = pi (r)^2 so the area of a semicircle is found with the equation

A = pi(r)^2 / 2

Since the diameter is 8m then the radius is 8/2 = 4m

Let's plug it into the equation

A = pi(r)^2 / 2
A = 3.14(4)^2 / 2
A = 25.12m^2

Now let's subtract the area of the semicircle from the area of the rectangle and we will be left with the shaded area.

96m^2 - 25.12m^2 =

70.88m^2

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25 POINTS!

WITCHER [35]

Answer:

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The sine of the angle is: positive

Step-by-step explanation:

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2 years ago

What is the solution to the equation 6y –2(y + 1) = 3(y – 2) + 6?

AleksAgata [21]

Answer: y=2

Step-by-step explanation:

Distribute the numbers -2 and 3...

6y-2y-2=3y-6+6

The -6 and 6 cancel each other out...

6y-2y-2=3y

Combine like terms....

4y-2=3y

Move 4y over....

-2=-y

Multiply both sides by -1.....

2=y

That’s your solution! Hope this helps!

8 0

3 years ago

A right triangle has a hypotenuse of 70 feet and a leg of 35 feet. What is the length of the other leg? Select one:

saveliy_v [14]

H = 70ft.
P = 35ft.
B = ?

H^2 = P^2 + B^2
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B^2 = 4900-1225
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3 0

2 years ago

The archway of the main entrance of a university is modeled by the quadratic equation

Scilla [17]

Answer:

B. (1,5) and (5.25, 3.94)

Step-by-step explanation:

The answer is where the 2 equations intersect.

We need to solve the following system of equations:

y = -x^2 + 6x

4y = 21 - x

From the second equation:

x = 21 - 4y

Plug this into the first equation:

y = -(21 - 4y)^2 + 6(21 - 4y)

y = -(441 - 168y + 16y^2)+ 126 - 24y

y = -441 + 168y - 16y^2 + 126 - 24y

16y^2 + y - 168y + 24y + 441 - 126 = 0

16y^2 - 143y + 315 = 0

y = [-(-143) +/- sqrt ((-143)^2 - 4 * 16 * 315)]/ (2*16)

y = 5, 3.938

When y = 5:

x = 21 - 4(5) = 1

When y = 3.938

x = 21 - 4(3.938) = 5.25.

3 0

2 years ago

Giá trị x thoả mãn x(x − 3) + x − 3 =0 là

svet-max [94.6K]

Here we will use the trial and error method.

We will try putting different values of x.

$\begin{gathered}\\ \sf\longmapsto 1(1-3)+1-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto (-2)+1-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -2+1-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -2-2=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -4≠0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 2(2-3)+2-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 2(-1)+2-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -2+2-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -3≠0\end{gathered}$

<h2>Again,</h2><h3>x=3</h3>

$\begin{gathered}\\ \sf\longmapsto 3(3-3)+3-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 3(3-3)=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 3×0=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 0=0\end{gathered}$

<h3>☣Hence, The value of X as 3 satisfies the equation!</h3>

5 0

2 years ago