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Oliga [24]
3 years ago
6

Does anyone know this?

Mathematics
1 answer:
Ierofanga [76]3 years ago
7 0
61.51.

You can get this by using the tangent function and solving for the opposite side. It will ultimately give you the value of 30Tan(64), which rounds to the answer above. 
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Answer:

-5 (3b+4)

Step-by-step explanation:

-15b - 20

-5 (3b+4)

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F(x) = x2 – 3x – 2 is shifted 4 units left. The result is g(x). What is g(x)?
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\bf \qquad \qquad \qquad \qquad \textit{function transformations}&#10;\\ \quad \\\\&#10;&#10;\begin{array}{rllll} &#10;% left side templates&#10;f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}&#10;\end{array}

\bf \begin{array}{llll}&#10;% right side info&#10;\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}&#10;\\\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\end{array}

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now... your expression is not in vertex form, that's ok... to do a horizontal shift to the left by 4 units, we can simply, add the C  and B component to the "x" variable, C=4, B =1, that way the horizontal shift of C/B or 4/1 is just +4, giving us a horizontal shift to the left

\bf f(x)=x^2-3x-2\impliedby \textit{let's change that for }f(1x+4)&#10;\\\\\\&#10;f(1x+4)=(1x+4)^2-3(1x+4)-2&#10;\\\\\\&#10;f(x+4)=x^2+8x+16-3x-12-2&#10;\\\\\\&#10;f(x+4)=x^2-5x+2\impliedby g(x)


8 0
3 years ago
Read 2 more answers
what are the hypothesis and conclusion of the following statement write it as a conditional residents of key west live in florid
vladimir1956 [14]
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3 years ago
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Anuta_ua [19.1K]

Answer:

<h2>x = -2 or x = 2</h2>

Step-by-step explanation:

Domain:\\8+x\neq0\ \wedge\ 8-x\neq0\\\boxed{D:x\neq-8\ \wedge\ x\neq8}\\\\\dfrac{60}{8+x}+\dfrac{60}{8-x}=16\\\\\dfrac{60(8-x)}{(8+x)(8-x)}+\dfrac{60(8+x)}{(8+x)(8-x)}=16\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\\dfrac{60(8-x)+60(8+x)}{8^2-x^2}=16\qquad\text{use the distributive property}\\\\\dfrac{(60)(8)+(60)(-x)+(60)(8)+(60)(x)}{64-x^2}=16\\\\\dfrac{480-60x+480+60x}{64-x^2}=16\qquad\text{combine like terms}\\\\\dfrac{(480+480)+(-60x+60x)}{64-x^2}=16\\\\\dfrac{960}{64-x^2}=16

\dfrac{960}{64-x^2}=\dfrac{16}{1}\qquad\text{cross multiply}\\\\16(64-x^2)=(960)(1)\qquad\text{use the distributive property}\\\\(16)(64)+(16)(-x^2)=960\\\\1024-16x^2=960\qquad\text{subtract 1024 from both sides}\\\\1024-1024-16x^2=960-1024\\\\-16x^2=-64\qquad\text{divide both sides by (-16)}\\\\\dfrac{-16x^2}{-16}=\dfrac{-64}{-16}\\\\x^2=4\to x=\pm\sqrt4\\\\x=\pm2\in D

3 0
4 years ago
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