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2 years ago

I need help in 2 3 4 5 6 7 8

Mathematics

1 answer:

ahrayia [7]2 years ago

7 0

Yo i can't see wut it says

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Find the area of the shaded region

ale4655 [162]

Answer:

379.56ft²

Step-by-step explanation:

Oh lordy why, okay we got this:

So we're going to start with the area of the square, this is easy because all of the sides are 25ft, so we just have to plug into our formula for squares (base x height):

25 x 25 = 625

Now we have to do the semicircle, remember that the formula for the area of a circle is pi x radius², we're using 3.14 for pi, so we have to find the radius (half the diameter):

25 ÷ 2 = 12.5

12.5² = 156.25

3.14 x 156.25 = 490.625‬

Normally we would stop here, but because this is only a semicircle, we have to divide by two:

490.625‬ ÷ 2 = 245.3125‬

So, now that we have our two areas, we have to subtract the semicircle from the square:

625 - 245.3125‬ = 379.6875‬

Now, I realize that his answer isn't in your cute little word bank, but that because I rounded pi, so the closest is 379.56

So, the area of the figure is 379.56ft²

if this is wrong let me know so I can fix it

hope this helps:)

5 0

2 years ago

Can somebody help me

Alex

31 i hope u got it right !!

3 0

2 years ago

Betty is having invitations printed for a holiday party. It costs $3.70 to set up the printing machine and $1.55 per invitation.

PIT_PIT [208]

Y = 1.55x + 3.70
1.55 x 150 = 232.5
232.5 + 3.70 = 236.2

The answer is C 236.20 hope that helped.

4 0

2 years ago

MATH URGENT

maks197457 [2]

${x}^{2} + 4x - 12 = 0 \\ {x}^{2} + 4x + 4 - 4 - 12 = 0\\ {(x + 2)}^{2} - 16 = 0 \\ x = - + \sqrt{16} - 2 \\ x = + \: or - 4 - 2 \\ x = - 6 \: \: x = 2$
- subtract the 12 from both sides so that it becomes the last constant term in the quadratic equation which should now equal 0.
- take the 4x
- half the coefficient of 4 (2)
- square it (4)
- add it to the equation (+4)
- subtract it from the equation (-4)
- factorise the square (x+2)^2 expands to (x^2 + 4x + 2) as {a+b}^2={a^2 + ab + ba + b^2}
- now the equation is in turning point form.
- to find x, add 16 and square root 16 and (x+2)
- subtract 2 from positive or negative 4 (as -4^2 and 4^2 both equal 16).
- This should give you two values for x, -6 and 2.

I really hope that this helped :)

7 0

2 years ago

If cos theta= -8/17 and theta is in quadrant 3, what is cos2 theta and tan2 theta

Karo-lina-s [1.5K]

$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad 
\begin{array}{llll}
\textit{now, hypotenuse is always positive}\\
\textit{since it's just the radius}
\end{array}
\\\\\\
thus\qquad cos(\theta)=\cfrac{-8}{17}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}$

since the hypotenuse is just the radius unit, is never negative, so the - in front of 8/17 is likely the numerator's, or the adjacent's side

now, let us use the pythagorean theorem, to find the opposite side, or "b"

$\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite
\end{cases}
\\\\\\
\pm\sqrt{17^2-(-8)^2}=b\implies \pm\sqrt{225}=b\implies \pm 15=b$

so... which is it then? +15 or -15? since the root gives us both, well
angle θ, we know is on the 3rd quadrant, on the 3rd quadrant, both, the adjacent(x) and the opposite(y) sides are negative, that means, -15 = b

so, now we know, a = -8, b = -15, and c = 17
let us plug those fellows in the double-angle identities then

$\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
\boxed{1-2sin^2(\theta)}\\
2cos^2(\theta)-1
\end{cases}
\\ \quad \\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
-----------------------------\\\\
cos(2\theta)=1-2sin^2(\theta)\implies cos(2\theta)=1-2\left( \cfrac{-15}{17} \right)^2
\\\\\\
cos(2\theta)=1-\cfrac{450}{289}\implies cos(2\theta)=-\cfrac{161}{289}$

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\left( \frac{-15}{-8} \right)}{1-\left( \frac{-15}{-8} \right)^2}
\\\\\\
tan(2\theta)=\cfrac{\frac{15}{4}}{1-\frac{225}{64}}\implies tan(2\theta)=\cfrac{\frac{15}{4}}{-\frac{161}{64}}
\\\\\\
tan(2\theta)=\cfrac{15}{4}\cdot \cfrac{-64}{161}\implies tan(2\theta)=-\cfrac{240}{161}$

6 0

2 years ago