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jeka94
4 years ago
6

I need 27 50 of fuel in %

Mathematics
1 answer:
barxatty [35]4 years ago
8 0

Answer:

54%

Step-by-step explanation:

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The expression m÷4 is the distance each person runs in a relay race that is m miles. How far does each person run in a relay rac
inna [77]

Answer:

5

Step-by-step explanation:

Replace 20 instead of m

20/4=5

5 0
3 years ago
Read 2 more answers
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

7 0
3 years ago
Martha drew a pair of intersecting non-perpendicular lines, I and m. She numbered one pair of vertical angles <1 and <2, a
RUDIKE [14]

Answer:

  • D. Because <1 and <2 are each supplementary to <3, they are therefore congruent

Step-by-step explanation:

A. m<1 + m<2 + m<3 + m<4 = 360

  • Incorrect in terms of the proof

B. The sum of the measures of angles of a triangle is 180°

  • Not relevant

C. The measure of all right is 90°

  • Not relevant

D. Because <1 and <2 are each supplementary to <3, they are therefore congruent

  • Correct
8 0
3 years ago
If two angles are both obtuse, the two angles are equal
yan [13]
If two angles are equal, then the two angles are obtuse.

5 0
4 years ago
Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 &lt; 2pi. Show work please!
yan [13]

Answer:

\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}

Step-by-step explanation:

You need 2 things in order to solve this equation:  a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

cos(2\theta)=1-2sin^2(\theta)

so we will make that replacement, getting everything in terms of sin:

sin(\theta)+1=1-2sin^2(\theta)

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

2sin^2(\theta)+sin(\theta)=0

We can factor out the sin(theta), since it's common in both terms:

sin(\theta)(2sin(\theta)+1)=0

Because of the Zero Product Property, either

sin(\theta)=0 or

2sin(\theta)+1=0

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi.  They are:

\theta=\frac{\pi}{2},\frac{3\pi}{2}

The next equation needs to first be solved for sin(theta):

2sin(\theta)+1=0 so

2sin(\theta)=-1 and

sin(\theta)=-\frac{1}{2}

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval.  They are:

\theta=\frac{2\pi}{3},\frac{4\pi}{3}

7 0
3 years ago
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