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Kipish [7]
3 years ago
10

Answer: ______pencils,_____ rulers.

Mathematics
1 answer:
Paul [167]3 years ago
4 0
15 pencils - £1.20
£3.80 left
12 rulers - £3.60
20p left
2 pencils - 16p

17 pencils, 12 rulers
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KatRina [158]

Answer: floating variable.

Step-by-step explanation:

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3 years ago
I know it has to be a negative...........
sleet_krkn [62]

Answer:

m=-3

Step-by-step explanation:

  1. Choose 2 points on the graph. (-2,8) (3,-7)
  2. Apply slope formula.

m=\frac{rise}{run} =\frac{y2-y1}{x2-x1} \\\\\frac{-7-(8)}{3-(-2)} = \frac{-15}{3} =-3\\\\m=-3

I hope I choose the right cordinates...

4 0
4 years ago
What’s the unit rate of 16/32
Gala2k [10]

0.5 unit per unit (or any other label)

4 0
3 years ago
Read 2 more answers
The expression 1.5*1.09 models the housing costs, in thousands of dollars, for summer term t years since Chinedu applied to his
Alex777 [14]

Answer: Multiplicative rate of change in housing costs, for summer term t years since Chinedu applied to his college.

Step-by-step explanation:

Given: 1.5\times1.09^t models the housing costs, in thousands of dollars, for summer term t years since Chinedu applied to his college.

Since, the above expression is an exponential function. [ t is the exponent ]

The standard exponential function is given by :-

y=Ab^x, A is the initial amount , b is the multiplicative rate of change with respect to time period x.

as compare to the given expression, we get b= 1.09, which is the multiplicative rate of change in the housing costs, for summer term t years since Chinedu applied to his college.

4 0
3 years ago
What sine function represents an amplitude of 1, a period of 2π, a horizontal shift of π, and a vertical shift of −4?
Gwar [14]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\\\
f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}
\end{array}
\\\\
-------------------\\\\

\bf % template detailing
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\

\bf \bullet \textit{vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\left. \qquad  \right. \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\left. \qquad  \right. \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)

with that template in mind, let's see

the period of 2π, well, for that, you have to do nothing, because that's sine original period

Amplitude of 1, well, that's also sine's original amplitude

horizontal shift/phase... ok, that means C/B = π, now, it doesn't say if to the left or the right, so I gather either is ok, so...let's do the right

you could use then, C = -π and B = 1, that gives you -π/1 or -π

vertical shift of -4, well, that simply means D = -4

f(θ) = sin(θ - π) - 4   <-- will do
5 0
4 years ago
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