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3 years ago

Davon marks a 4-inch paper strip into equal parts as shown below. Label the whole and quarter inches on the paper strip.

Mathematics

1 answer:

stiks02 [169]3 years ago

3 0

Answer:

Step-by-step explanation:

Divide the length of the paper strip into four equal sections of 1 inch each.

Mark the left end of the strip '0' and the right end '4.' Label the '1,' '2,' '3' points.

Between these marks, label the strip as follows:

'1/4,' '1/2,' '3/4,' 1 (already marked), '1 1/4,' '1 1/2,' and so on.

Next time, please be sure to share ALL parts of the question with which you want help. Thank you.

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What is 3/2m-m=4+1/2m

Annette [7]

Answer:

answer is 08

Step-by-step explanation:

3/2m-m=4+1/2*m

3m/2-m=4+1/2*m

3m/2-m=4+1/2m

3m/2-m=4+1m/2

3m/2-m=4+m/2

2(3m/2-m)=2(4+m/2)

m=m+8

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3 years ago

For all functions of the form f(x) = ax2 + bx + c, which is true when b = 0?

Klio2033 [76]

When b=0
f(x)=ax^2+c

test each
A. with x^2-1, A is false
B. with -x^2-1, B is false
C. cannot find contradiction
D. the axis is actually x=0, 0 is nithere positive nor negative, false

answer is C

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4 years ago

Read 2 more answers

In a class 32 students, 4students we're homesick with the flu on Thursday. what percentage of the students were absent on Thursd

Fynjy0 [20]

Answer:

12.5%

Step-by-step explanation:

4 * 100 / 32 = 12.5

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3 years ago

use the guide to construct a two column proof proving △aec≅△deb, given that ca is parallel to db and e is the midpoint of ad. gi

MrMuchimi

By using the AAA congruence property of Triangles, △aec≅△deb is proved

It is given two triangles, Δaec and Δdeb, where e is the common point known as the midpoint of ad

Also, it is given that,

ca ║db

We need to prove that, △aec≅△deb

Then we'll use AAA congruence property of Triangles to prove the situation

As ca ║db

then, ∠cae = ∠ebd (Alternate angles)

∠ace = ∠edb (Alternate angles)

and ∠aec = ∠deb (common angles)

Thus, by AAA congruence property, △aec≅△deb

Hence, proved

To learn more about, congruence property, here

brainly.com/question/2039214

#SPJ4

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1 year ago

How does the graph of f(x)=3lx+2l+4 relate to its parent function?

Sergeeva-Olga [200]

$\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, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below

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3 years ago