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Afina-wow [57]
3 years ago
7

So uh can someone answer this for me please i don’t understand it

Mathematics
2 answers:
lisabon 2012 [21]3 years ago
8 0

so the area = 36

which means that (2.5x + 1.5) X 4 = 36

so then you need to cancel out things on the left, by that I mean, to get rid of the x4 you must ÷4 on each side

2.5x + 1.5= 9

2.5x= 7.5

x= 3

Hope that helps

Simora [160]3 years ago
4 0

Answer:

jjjjjiii

5

Step-by-step explanation:

2.5x+1.5 × 4 = 36

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Type the correct answer for each.
xxTIMURxx [149]
If point \left(x,\ \frac{\sqrt{7}}{3}\right) is on the unit ircle, then:

x^2+\left( \frac{ \sqrt{7} }{3} \right)^2=1 \\  \\ \Rightarrow x^2+ \frac{7}{9} =1 \\  \\ \Rightarrow x^2=1- \frac{7}{9} = \frac{2}{9}  \\  \\ \Rightarrow x= \sqrt{ \frac{2}{9} } = \frac{ \sqrt{2} }{3}

Since, the point is in the second quadrant, x is negative.

Thus, (x,\ y)=\left(x,\ \frac{\sqrt{7}}{3}\right)=\left(-\frac{ \sqrt{2} }{3},\ \frac{\sqrt{7}}{3}\right)

Part A:

3 \sqrt{7} =6\left( \frac{ \sqrt{7} }{2} \right) \\  \\ =6\left( \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \right) \\  \\ =6\left( \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } \right)=-6\left( \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } \right) \\  \\ =-6\left( \frac{y}{x} \right)=-6\tan\theta

Therefore, -6\tan\theta=3\sqrt{7}.



Part B:

- \frac{ \sqrt{7} }{2} =- \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} }  \\ \\ =- \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } = \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } = \frac{y}{x}  \\  \\ =\tan\theta

Therefore, \tan\theta=- \frac{ \sqrt{7} }{2}
8 0
3 years ago
F(x) = 3x + x3
____ [38]

Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 3x + x³

Taking differentiate

\frac{d}{dx}\left(3x+x^3\right)

\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'

=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)

solving

\frac{d}{dx}\left(3x\right)

\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'

=3\frac{d}{dx}\left(x\right)

\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1

=3\cdot \:1

=3

now solving

\frac{d}{dx}\left(x^3\right)

\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}

=3x^{3-1}

=3x^2

Thus, the expression becomes

\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)

                    =3+3x^2

Thus,

f'(x) = 3 + 3x²

Given that f'(x) = 15

substituting the value  f'(x) = 15 in f'(x) = 3 + 3x²

f'(x) = 3 + 3x²

15 =  3 + 3x²

switch sides

3 + 3x² = 15

3x² = 15-3

3x² = 12

Divide both sides by 3

x² = 4

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{4},\:x=-\sqrt{4}

x=2,\:x=-2

Thus, the value of x​ will be:

x=2,\:x=-2

5 0
2 years ago
85,146 rounded to the nearest thousand
larisa [96]
85,000 is your answer because The one in 146 is less then five
4 0
3 years ago
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Amira is painting a rectangular banner 2 1/4 yards wide in the cafeteria. The banner will have a blue background. Amira has enou
elena55 [62]
Area of a rectangle:
A = w x h, where w is the width and h is the height.
w = 2.25 yd
A = 1.5 yd²
1.5 = 2.25 · h
h = 1.5 : 2.25 = 0.6667 = 2/3 yd
Answer:
The height of a banner is 2/3 yard.
4 0
3 years ago
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A polyhedron with 16 edges cannot be a prism. Explain why this must be true
Readme [11.4K]

Answer:

This prism has: 5 Faces. 6 Vertices (corner points) 9 Edges.

Step-by-step explanation:

3 0
2 years ago
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