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erma4kov [3.2K]
4 years ago
5

How would I do this confused

Mathematics
2 answers:
Anna35 [415]4 years ago
7 0
Multiply 1/4 by 2 to get 2/8. Don't forget to add 2 to your 2/8. That means that 2 2/8 minus 3/8 equals 1 7/8
nikklg [1K]4 years ago
4 0

-3/8+ 2 1/4 = 1.875 because it doesn't always stay a Fraction!!!

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Germany Ron 7 miles per week and increase his distances by 1 mile in each week tell tony runs 3 miles per week and increases his
Lelechka [254]
X represents the amount of weeks:
7+x for Germany and 3+2x for Tony they have to be equal for them to run the same distance:
3+2x=7+x
-3 from each side:
2x=4+x
-x from each side:
x=4
They run the same distance on the fourth week, that distance is 7+4=11

Hope this helps :)
4 0
3 years ago
What is the lateral surface area of a cube with side length 9 cm?
ycow [4]

Answer:

A

Step-by-step explanation:

Quizlet

8 0
3 years ago
What is the period and midline?
OverLord2011 [107]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\

\end{array}\qquad

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\
\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}\\
\bullet \textit{vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\


\end{array}

\bf \begin{array}{llll}
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}


so if you notice yours \bf \begin{array}{llll}
3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\
&\ \uparrow&\uparrow \\
&B&D 
\end{array}

now.. normally the function \bf 3.2cos&\left( \frac{5}{3}\theta \right)
 has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline  up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is 2\pi
so, the new period will be \bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)

6 0
3 years ago
Which ordered pair is a solution to this equation?
yarga [219]

Let's test each alternative :

2 x - 9 y = 38      ( 7,5 ) :

( 2 * 7 ) -  ( 9 * 5 ) = 38

14 - 45  = 38

-31 ≠ 38   (  wrong )


( 4 , - 1 ) :

( 2 * 4 ) - ( 9 * ( - 1 ) =  38

8 + 9  = 38

17 ≠ 38  ( wrong )

( 3 , - 3 ) :

2 * ( 3 ) - ( 9 * ( - 3 ) = 38

6 + 27 = 38

33 ≠ 38  ( wrong )

( 1 , - 4 ) :

2* ( 1 ) - 9 * ( - 4 ) = 38

2 + 36 = 38

38 = 38    ( Correct alternative )

( 1 , - 4 ) is correct alternative


3 0
3 years ago
F(x)= -5x+9 ; reflection in the y-axis a.g(x)=5x-9 b.g(x)=5x+9 c.g(x)=9x-5 d.g(x)=5x-5
Sunny_sXe [5.5K]
<h3>Answer: g(x) = 5x+9, second choice</h3>

Explanation:

When we reflect over the vertical y axis, every negative x value becomes positive and vice versa. So we effectively replace x with -x like so

f(x) = -5x+9

f(-x) = -5(-x)+9 ... replace every x with -x

f(-x) = 5x+9

g(x) = 5x+9

A point like (1,4) on f(x) will reflect over to (-1,4) on g(x). If we apply this to every point on f(x), then all of f(x) will reflect to g(x).

7 0
4 years ago
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