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timama [110]
2 years ago
11

The bases on a baseball diamond are 90 feet apart. how far is it from home plate to second base?

Mathematics
2 answers:
Karolina [17]2 years ago
3 0
See diagram

so we got a right triangle

so for a right triangle, when we've got legs a and b and hytponuse c then
a²+b²=c²


so the legs are 90 and 90
hytponuse is distance from home to 2nd

so
90²+90²=c²
8100+8100²=c²
16200=c²
sqrt both sides
90√2=c
aprox
127.279ft
about 127ft

Brilliant_brown [7]2 years ago
3 0
Assuming that the bases are all within right angles of each other:
Form a triangle with vertices at home base, first, and second. Home to first is 90ft and first to second is also 90ft.
Use the formula a^2+b^2=c^2 to find the the missing length.
90^2+90^2=c^2
8100+8100=c^2
16200=c^2
Sqrt(16200)=c
c=127.279
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What is parenthesis ?
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2 years ago
F(x) = x2 – 3x – 2 is shifted 4 units left. The result is g(x). What is g(x)?
Ray Of Light [21]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}

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

\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}
\end{array}

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)
\\\\\\
f(1x+4)=(1x+4)^2-3(1x+4)-2
\\\\\\
f(x+4)=x^2+8x+16-3x-12-2
\\\\\\
f(x+4)=x^2-5x+2\impliedby g(x)


8 0
2 years ago
Read 2 more answers
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Nikolay [14]

Answer:

13

Step-by-step explanation:

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5 0
3 years ago
PLEAS HELP ASAP :)))))
Jet001 [13]

Answer:

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Step-by-step explanation:

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8 0
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