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aleksklad [387]
2 years ago
13

H=-16t^2+96t+4solve by factoring​

Mathematics
2 answers:
BigorU [14]2 years ago
8 0

Answer:

t = 3 ± ½√37

Step-by-step explanation:

0 = -16t² + 96t + 4

This isn't factorable, so complete the square or use quadratic formula.  Completing the square:

16t² − 96t = 4

t² − 6t = 1/4

t² − 6t + 9 = 1/4 + 9

(t − 3)² = 37/4

t − 3 = ±½√37

t = 3 ± ½√37

zlopas [31]2 years ago
4 0

Answer:

6 plus or minus square root of 37/2

Step-by-step explanation:

Is the quadratic formula allowed or no?

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D

Step-by-step explanation:

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3 years ago
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PLEASE SOLVE AND CHECK. SHOW COMPLETE SOLUTION
Alex17521 [72]

<u>Solution</u><u>:</u>

\sqrt{4x + 13}  = x + 2

  • First square both sides.

=  > ( \sqrt{4x + 13} ) ^{2}  = (x + 2) ^{2}

  • Now, square root and square gets cancel out in the LHS. And in the RHS, apply the identity: (a + b)² = a² + 2ab + b².

=  > 4x + 13 =  {(x)}^{2}  + 2 \times x \times 2 + (2) ^{2} \\  =  > 4x + 13 =  {x}^{2}   + 4x + 4

  • Now, transpose 4x and 4 to LHS.

=  > 4x - 4x + 13 - 4 =  {x}^{2}  \\

  • Now, do the addition and subtraction.

=  >  {x}^{2}  = 9 \\  =  >  x =  \sqrt{9}  \\  =  > x = ±3

<u>Answer</u><u>:</u>

<u>x </u><u>=</u><u> </u><u>±</u><u> </u><u>3</u>

Hope you could understand.

If you have any query, feel free to ask.

3 0
2 years ago
What is the image point of (-1, 1) after a translation right 2 units and down 1 unit?
stich3 [128]

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Step-by-step explanation:

7 0
3 years ago
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Find thhe remainder when 7^203 is divided by 4
Aleksandr [31]
Using the square-and-multiply approach, we have

7^{203}=7\times(7^{101})^2
7^{101}=7\times(7^{50})^2
7^{50}=(7^{25})^2
7^{25}=7\times(7^{12})^2
7^{12}=(7^6)^2
7^6=(7^3)^2
7^3=7\times7^2

and so, using the property that, if a_1\equiv b_1\mod n and a_2\equiv b_2\mod n, then a_1a_2\equiv b_1b_2\mod n, we get

7\equiv3\mod4
7^2\equiv9\equiv1\mod4
7^3\equiv7\times1\equiv7\equiv3\mod4
7^6\equiv9\equiv1\mod4
7^{12}\equiv1\mod4
7^{25}\equiv7\times1\equiv7\equiv3\mod4
7^{50}\equiv9\equiv1\mod4
7^{101}\equiv7\times1\equiv7\equiv3\mod4
7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4
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3 years ago
Can somebody help me please.​
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The Y intercept is (0,275) and the X intercept is (125,0). the intercepts should be the point of when the line hits the axis'

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