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forsale [732]
3 years ago
12

What’s the value of x for a-d?

Mathematics
1 answer:
patriot [66]3 years ago
8 0

Answer:

a = 76 b = 22 c = 58 d = 68

Step-by-step explanation:

A. 180 - 133 = 47

so you have two angles found, which is 57 and 47. Add them together and subtract from 180. You get 76

B. 68 + 90 = 158

You subtract the 158 from 180 and get 22

C. This took some guess work, but when you multiply 58 by 2 you get 116. When subtracted from 180 you get 64. If you add 58 + 58 + 64, you get 180, so this made sense

D. There's an actual rule in trig that explains this, but I forgot it. If you look at the top angle, the other side of the line added to the 68 is a right angle. So you subtract 68 from 90 and get 12. Because there is already the right angle in the triangle, you just need the other two angles to equal 90, and because of the 68 that we found in the other triangle, this made sense.

Sorry, not the best at explanations. Hope that helped

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A recipe includes 2 cups of flour and two thirds
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The answer to your question is 2/3 because there are 2 cups and 2/3 of buttermilk meaning 33.3333.. so it would be 20/30 then 2/3
6 0
3 years ago
If the regular selling price of a silk scarf is $49 and the markdown rate is 38%, what is its sale price?
Aleks04 [339]

Answer:

$30.38

Step-by-step explanation:

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The selling price of the scarf will be 0.62(49) = 30.38

4 0
4 years ago
Linda goes water-skiing one sunny afternoon. After skiing for 15 min, she signals to the driver of the boat to take her back to
trasher [3.6K]
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2


abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.


d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
4 0
3 years ago
Simplify the number into simplest radical form. Use the factor tree to help determine the factors. StartRoot 96 Endroot StartRoo
NikAS [45]

Answer:

[tex]4608\sqrt{3}[/tex]

Step-by-step explanation:

1. \sqrt{96} *\sqrt{6}* 2*\sqrt{6}*4 *\sqrt{6} *4*\sqrt{3}

2. \sqrt{2^{5} *3} \sqrt{6} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}   <em>factoring 96</em>

<em>since \sqrt{2^{5}*3 } = \sqrt{2^{5} } \sqrt{3}</em>

3. \sqrt{2^{5} } \sqrt{3}\sqrt{6} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}

<em>using exponent rule - (a^{b}) ^{c} = a^{bc}</em>

<em> \sqrt{2^{5} } = 2^{5/2}</em>

4. 2^{5/2}\sqrt{3}\sqrt{2*3} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}

<em>doing some simple simplification and 4=2^{2}  and 6=2*3</em>

5. 2^{5/2} \sqrt{3} \sqrt{2} \sqrt{3} *2\sqrt{2} \sqrt{3} *2^{2}\sqrt{2} \sqrt{3}*4\sqrt{3}

<em>collecting the roots on one side and applying exponent rule</em>

6. \sqrt{3} \sqrt{3}\sqrt{3} \sqrt{3}\sqrt{2} \sqrt{2}\sqrt{2} *2^{5/2+1+2+2} \sqrt{3}

<em>Applying exponents rule on all \sqrt{3} and \sqrt{2}</em>

<em>7. 2^{1/2+1/2+1/2} *2^{5/2+1+2+2}*3^{1/2+1/2+1/2+1/2+1/2}</em>

<em>combining all powers of 2</em>

8. 2^{1/2+1/2+1/2+5/2+1+2+2}*3^{1/2+1/2+1/2+1/2+1/2}

<em>Simplifying</em>

9. 2^{9} *3^{2}\sqrt{3}

10. 512*9\sqrt{3}

11. 4608\sqrt{3}

3 0
3 years ago
Read 2 more answers
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