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Ludmilka [50]
2 years ago
6

A recipes calls for 1/4 cups of flour. Carter only has a measuring cup that holds 1/8 cup. how can carter measure the flour he n

eeds for his recipe?
Mathematics
2 answers:
Aloiza [94]2 years ago
6 0

Answer:

He can use the carter twice which holds 1/8 cups of flour because twice the 1/8 cups of flour can provide 1/4 cups of flour.

Step-by-step explanation:

Given :

A recipes need 1/4 cups of flour.

Carter  has a measuring cup that holds 1/8 cup.

To Find : How can carter measure the flour he needs for his recipe?

Solution :

Since he need 1/4 cups of flour .

And Carter can measure only 1/8 cup.

Since we can see that

2*\frac{1}{8} =\frac{1}{4}

⇒ He can use the carter twice which holds 1/8 cups of flour because twice the 1/8 cups of flour can provide 1/4 cups of flour.

Sati [7]2 years ago
4 0
Since 2/8 is equal to 1/4 you need 2 of the 1/8 cups.
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Answer:

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Step-by-step explanation:

There are 2000 items, so n = 2000

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2 years ago
Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g
SpyIntel [72]
\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, let's see    \bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad  B
\end{array}

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.
8 0
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Answer:

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3 years ago
Use the quadratic formula to solve for x. 4x2 +9x-1=0​
insens350 [35]

Answer:

x=\frac{-9\pm\sqrt{65} }{8}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Standard Form: ax² + bx + c = 0
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Step-by-step explanation:

<u>Step 1: Define</u>

4x² + 9x - 1 = 0

<u>Step 2: Define variables</u>

a = 4

b = 9

c = -1

<u>Step 3: Find roots</u>

  1. Substitute:                    x=\frac{-9\pm\sqrt{9^2-4(4)(-1)} }{2(4)}
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  3. Multiply:                        x=\frac{-9\pm\sqrt{81-16} }{8}
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