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Elodia [21]
3 years ago
10

Tina placed a 12 meter rope along one side of the bicycle path. She hung a ribbon on each end of the rope and every 3 meters in-

between. How many ribbons did she hang?
A. 4
B. 5
C. 6
D. 7
Mathematics
1 answer:
olga_2 [115]3 years ago
5 0

Answer:

4

Step-by-step explanation:

If the tape is 12m,

she hung the tape in between every 3m, and at both ends

0, 3, 6, 9, 12

these are the lengths she hung ribbons at

Total number of ribbons = 4

A is correct

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Hope this helps
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Given the equation y = 0.2 x 4^x , find x when y = 1000
Gnoma [55]

Answer:

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6 0
3 years ago
-12x - 12y = 4 3x + 3y = 0
Svetradugi [14.3K]

Answer:

no Solution

Step-by-step explanation:

-12x-12y=4\\ 3x+3y=0

12x-12y=4

add 12y to both sides

12x-12y+12y=4+12y

divid both sides by -12

\frac{-12x}{-12}=\frac{4}{-12}+\frac{12y}{-12}

simplfy

x=-\frac{1+3y}{3}

\mathrm{Substitute\:}x=-\frac{1+3y}{3}

\begin{bmatrix}3\left(-\frac{1+3y}{3}\right)+3y=0\end{bmatrix}

\begin{bmatrix}-1=0\end{bmatrix}

7 0
3 years ago
WILL MARK BRAINLIEST IF CORRECT!!!!
lozanna [386]

Answer:

yo mama

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Suppose f (x) = x^2. Find the graph of f(x+2)
NISA [10]
\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)={{  A}}({{  B}}x+{{  C}})+{{  D}}
\\\\
~~~~y={{  A}}({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)={{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\\\
f(x)={{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\\\
f(x)={{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\\\\
--------------------

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
~~~~~~\textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
~~~~~~if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
~~~~~~if\ {{  D}}\textit{ is negative, downwards}\\\\
~~~~~~if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind,

\bf f(x)=x^2\qquad \quad f(x+2)=(x+2)^2\implies f(x+2)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+2})^2\stackrel{D}{+0}

C = 2         B = 1        C/B = 2/1 or +2,    horizontal left shift of 2 units

f(x) shifted left by 2 units is f(x+2).
8 0
3 years ago
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