write a word problem that can be solve by drawing an array. Then draw the array and solve the problem.

Mathematics

1 answer:

Alborosie4 years ago

4 0

The answer is B hope this helped

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In the number 76,129 which digit has the least value

Andreas93 [3]

Number wise, 1.

If we were talking places, it would be the nine.

5 0

3 years ago

Read 2 more answers

3) Josie's collection of 39 coins includes twice as many nickels than quarters. If

Degger [83]

Answer:

13 quarters, 26 nickels.

Step-by-step explanation:

n will be number of nickels, q will be number of quarters, and we have 39 total coins

so n + q = 39 (number of nickels + number of quarters = 39 coins.

The problem also says there are twice as many nickels as quarters. so we can call the number of nickels n or we can call it 2q (2 times the number of quarters)

which means

n + q = 39

2q + q = 39

3q = 39

q = 13

and she has 2 times as many nickels as quarters... 2 x 13 = 26 nickels

LMK if you have questions.

If you want to double check the money,

n(0.05) + q(0.25) = 4.55 is the equation to use.

26(0.05) + 13(0.25) = 1.3 + 3.25 = $4.55

5 0

3 years ago

Graph the following sequence of transformations for

Sedbober [7]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------$

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

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

with that template in mind.

$\bf f(x)=\sqrt[3]{x}\qquad \boxed{-f(x+2)+4}\implies f(x)=\stackrel{A}{-1}\sqrt[3]{\stackrel{B}{1}x\stackrel{C}{+2}}\stackrel{D}{+4}\qquad 
\\\\\\
f(x)=-\sqrt[3]{x+2}+4$

A = -1, reflected over the x-axis

B = 1, C = 2, horizontal shift of C/B or 2, to the left.

D = 4, shifted upwards by 4 units.

check the picture below.