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Harrizon [31]
3 years ago
13

Help pls yall i will give points

Mathematics
1 answer:
katen-ka-za [31]3 years ago
8 0

Answer:

im gonna tell you its none promise

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True

Step-by-step explanation:

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In exponential growth functions, the base of the exponent must be greater than 1. How would the function change if the base of t
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If the base were 1, then the final value would never change...it would be a constant...
If the base is between zero and one, it is an exponential decay equation, so the final value would continually get smaller and smaller...
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There are 36 students in the class 1/6 students are wearing blue shirts and 2/3 are wearing white shirts how many are wearing a
Marat540 [252]

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The answer would be 1/6 kids are wearing shirts that are neither blue or white.

Step-by-step explanation:

So, to make this easier, lets convert 2/3 to a same denominator fraction.

We must ask, how many times does 3 go into six?

We multiply 3 by 2 and get six. We then multiply the numerator of 2 by the same, getting 4.

Our new fraction is 4/6 which is equal to 2/3.

Then we add 4 to 1, making it 5. That is the amount of students wearing blue and white shirts, which means

1/6 of the kids in the class are wearing shirts that are neither white nor blue.

Hope this helps,

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3 0
3 years ago
Factor the expression using GCF 16x+56y
koban [17]

Answer:

4(3x+14y)

Step-by-step explanation:

12x +56y

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4 0
3 years ago
Solve by quadratic equation​
Ymorist [56]
<h2>Question :</h2>

  • \tt \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}

<h2>Answer :</h2>

  • \large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}

<h2>Explanation :</h2>

\tt : \implies \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}

\tt : \implies \dfrac{(x+2)(x+2) + (x-2)(x-2)}{(x-2)(x+2)} = \dfrac{5}{6}

\tt : \implies \dfrac{(x+2)^{2} + (x-2)^{2}}{(x-2)(x+2)} = \dfrac{5}{6}

<u>Now, we know that</u> :

  • \large \underline{\boxed{\bf{(a+b)^{2} = a^{2} + b^{2}+ 2ab}}}
  • \large \underline{\boxed{\bf{(a-b)^{2} = a^{2} + b^{2} - 2ab}}}
  • \large \underline{\boxed{\bf{(a+b)(a-b) = a^{2} - b^{2}}}}

\tt : \implies \dfrac{x^{2}+2^{2}+ 2 \times x \times 2 + x^{2}+2^{2} - 2 \times x \times 2 }{x^{2}-2^{2}} = \dfrac{5}{6}

\tt : \implies \dfrac{x^{2}+ 4 + \cancel{4x} + x^{2}+ 4 - \cancel{4x}}{x^{2}-4} = \dfrac{5}{6}

\tt : \implies \dfrac{x^{2} + x^{2} + 4 + 4}{x^{2}-4} = \dfrac{5}{6}

\tt : \implies \dfrac{2x^{2} + 8}{x^{2}-4} = \dfrac{5}{6}

<u>By cross multiply</u> :

\tt : \implies (2x^{2} + 8)6= 5(x^{2}-4)

\tt : \implies 12x^{2} + 48 = 5x^{2}-20

\tt : \implies 12x^{2} + 48 - 5x^{2} + 20 = 0

\tt : \implies 7x^{2} + 68 = 0

\tt : \implies 7x^{2} + 0x + 68 = 0

<u>Now, by comparing with ax² + bx + c = 0, we have</u> :

  • a = 7
  • b = 0
  • c = 68

<u>By using quadratic formula</u> :

\large \underline{\boxed{\bf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}}}

\tt : \implies x = \dfrac{-(0) \pm \sqrt{(0)^{2} - 4(7)(68)}}{2(7)}

\tt : \implies x = \dfrac{0 \pm \sqrt{0 - 1904}}{14}

\tt : \implies x = \dfrac{\pm \sqrt{- 1904}}{14}

\tt : \implies x = \dfrac{\pm \sqrt{2\times 2\times 2\times 2\times 7\times 17}}{14}

\tt : \implies x = \dfrac{\pm \cancel{2} \times 2\sqrt{7\times 17}}{\cancel{14}}

\tt : \implies x = \dfrac{\pm2\sqrt{119}}{7}

\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}

Hence value of \bf x =\dfrac{\pm 2\sqrt{119}}{7}

5 0
3 years ago
Read 2 more answers
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