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Mumz [18]
3 years ago
14

Amy needs 6/8 gallon of fruit juice to make punch she needs an equal amount of sparkling water how much sparkling water does she

need
Mathematics
1 answer:
Roman55 [17]3 years ago
5 0
Amy would need 6/8 gallon of sparkling water, or 3/4 gallon, because 6/8=6/8 or 3/4.
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Complete the point-slope equation of the line through (-4,8) and (4,4)
S_A_V [24]

Answer:

y-4= -1/2(x-4)

Step-by-step explanation:

First you have to find the slope to write it in point slope

to do this you find change in Y over change in X

From -4 to 4 is postivite 8 and from 8 to 4 its -4

Now you do -4 divided by 8 and get -0.5 or -1/2 for slope now pick one of the points and put it in the formula y-y1=m(x-x1)

3 0
3 years ago
Read 2 more answers
A solution of salt and water is 20% salt. If 25% of the water is removed, what is the percentage of salt in the new solution?
Ratling [72]
So lets say that this salt and water solution consists of 100 liters.

Then that means 20 liters is salt, and 80 liters out of the 100 is water.



Now, we remove 25% of the water. 25% of 80 liters is 20 liters. So, we are left with 80-20, or 60 liters.



Now in all we have 20 liters of salt, and 60 liters of water.

The percentage of salt in the new mixture will be 20/(20+60) = 20/80 = 25%.



Your answer is 25%
4 0
3 years ago
3. Which equation is written in slope-intercept form?
NARA [144]
Y=mx+b is slope intercept form. it is y=3x+2 where 3/1 is the slope and 2 is the y-intercept.
5 0
3 years ago
The scale on a map is 2 cm= 5.5 miles. If two cities are 15 miles apart, how many cm apart would they be on the map?
xxTIMURxx [149]

Answer:

3 cm

Step-by-step explanation:

5 0
3 years ago
Find the average rate of change of the function over the given interval
sattari [20]
\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\

\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{4}\\
t_2=\frac{3\pi }{4}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{4} \right)-h\left( \frac{\pi }{4} \right)}{\frac{3\pi }{4}-\frac{\pi }{4}}
\\\\\\

\bf \cfrac{\frac{cos\left( \frac{3\pi }{4} \right)}{sin\left( \frac{3\pi }{4} \right)}-\frac{cos\left( \frac{\pi }{4} \right)}{sin\left( \frac{\pi }{4} \right)}}{\frac{\pi }{2}}\implies \cfrac{-1-1}{\frac{\pi }{2}}\implies \cfrac{-2}{\frac{\pi }{2}}\implies -\cfrac{4}{\pi }\\\\\\
-------------------------------\\\\

\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{3}\\
t_2=\frac{3\pi }{2}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{2} \right)-h\left( \frac{\pi }{3} \right)}{\frac{3\pi }{2}-\frac{\pi }{3}}
\\\\\\

\bf \cfrac{\frac{cos\left( \frac{3\pi }{2} \right)}{sin\left( \frac{3\pi }{2} \right)}-\frac{cos\left( \frac{\pi }{3} \right)}{sin\left( \frac{\pi }{3} \right)}}{\frac{9\pi -2\pi  }{6}}\implies \cfrac{\frac{0}{-1}-\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}{\frac{7\pi }{6}}\implies\cfrac{-\frac{1}{\sqrt{3}}}{\frac{7\pi }{6}}\implies -\cfrac{\sqrt{3}}{3}\cdot \cfrac{6}{7\pi }
\\\\\\
-\cfrac{2\sqrt{3}}{7\pi }
8 0
3 years ago
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