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kobusy [5.1K]
3 years ago
9

Help me with this please!!

Mathematics
2 answers:
enyata [817]3 years ago
7 0

Answer:

4

Step-by-step explanation:

3x + 8 = 5x ( inscribed angles on same arc are equal)

5x - 3x = 8

2x = 8

x = 8/ 2

x = 4

Hope it will help :)❤

Sedbober [7]3 years ago
4 0

Answer:

x = 4

Step-by-step explanation:

Since both of the angles shown in this picture cover the same arc, they have the same value, or:

3x + 8 = 5x

Solving this equation gives the answer:

2x = 8

x = 4

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Please help tell me what i put on estimate and product
ololo11 [35]

Answer:

You should have it right I checked the math the product is right and the estimate would mate sense so that should be right

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
"We might think that a ball that is dropped from a height of 15 feet and rebounds to a height 7/8 of its previous height at each
tatyana61 [14]

Answer:

Total Time = 4.51 s

Step-by-step explanation:

Solution:

- It firstly asks you to prove that that statement is true. To prove it, we will need a little bit of kinematics:

                             y = v_o*t + 0.5*a*t^2

Where,   v_o : Initial velocity = 0 ... dropped

              a: Acceleration due to gravity = 32 ft / s^2

              y = h ( Initial height )

                             h = 0 + 0.5*32*t^2

                             t^2 = 2*h / 32

                             t = 0.25*√h   ...... Proven

- We know that ball rebounds back to 7/8 of its previous height h. So we will calculate times for each bounce:

1st : 0.25*\sqrt{15}\\\\2nd: 0.25*\sqrt{15} + 0.25*\sqrt{15*\frac{7}{8} } + 0.25*\sqrt{15*\frac{7}{8} } = 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} }\\\\3rd: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 2*0.25*\sqrt{15*(\frac{7}{8} })^2\\\\= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2\\\\4th: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 2*0.25*\sqrt{15*(\frac{7}{8} })^3 \\\\

= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 0.5*\sqrt{15*(\frac{7}{8} })^3

- How long it has been bouncing at nth bounce, we will look at the pattern between 1st, 2nd and 3rd and 4th bounce times calculated above. We see it follows a geometric series with formula:

  Total Time ( nth bounce ) = Sum to nth ( \frac{1}{2}*\sqrt{15*(\frac{7}{8})^(^i^-^1^) }  - \frac{1}{4}*\sqrt{15})

- The formula for sum to infinity for geometric progression is:

                                   S∞ = a / 1 - r

Where, a = 15 , r = ( 7 / 8 )

                                   S∞ = 15 / 1 - (7/8) = 15 / (1/8)

                                   S∞ = 120

- Then we have:

                                  Total Time = 0.5*√S∞ - 0.25*√15

                                  Total Time = 0.5*√120 - 0.25*√15

                                  Total Time = 4.51 s

5 0
4 years ago
Write a word name as you would on a check for the dollar amount $681.66. Choose the word name that would be written on a check f
Crank
Are you sure this is college mathimatics bec this doesn't look like it
7 0
4 years ago
Please check my homework on Negative Indices!
rusak2 [61]

Our first expression is (2p^{-3} q^{8} )^{4}. Upon distributing the exponent 4 on all the terms, we get:

(2p^{-3} q^{8} )^{4}=2^{4}(p^{-3})^{4}(q^{8})^{4}=16p^{-12}q^{32}=\frac{16q^{32}}{p^{12}}

Therefore, your answer is correct for this part. :)

Second expression is (2m^{-4}n^{5})^{-2}. Upon distributing the exponent -2 on all the terms, we get:

(2m^{-4}n^{5})^{-2}=2^{-2}(m^{-4})^{-2}(n^{5})^{-2}=2^{-2}m^{8}n^{-10}=\frac{m^{8}}{4n^{10}}

Your second answer is correct too.

Our third expression is 2(5g^{-4}h^{-6})^{2}. Upon distributing the exponent 2 on all the terms, we get:

2(5g^{-4}h^{-6})^{2}=2(5^{2})(g^{-4})^{2}(h^{-6})^{2}=2(25)g^{-8}h^{-12}=\frac{50}{g^{8}h^{12}}

This one is not correct. Your answer would have been correct, if the exponent were -2 instead of 2 in this part.

Our forth and last expression is 4(2c^{-3}d^{6})^{-5}. Upon distributing the exponent -5 on all the terms inside the parenthesis, we get:

4(2c^{-3}d^{6})^{-5}=4(2^{-5})(c^{-3})^{-5}(d^{6})^{-5}=\frac{4}{2^{5}}(c^{15})(d^{-30})=\frac{4c^{15}}{32d^{30}}=\frac{c^{15}}{8d^{30}}

Therefore, your answer for this part is also correct.

Looking at your work, I don't think you made a mistake in number 3 also, probably mis-typed the question while writing here :)

3 0
4 years ago
Which table represents the graph
olya-2409 [2.1K]

I believe it is A.  I did this awhile ago.

6 0
3 years ago
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