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kaheart [24]
4 years ago
11

Is 1:4 and 6:18 proportional?

Mathematics
1 answer:
Alexxandr [17]4 years ago
5 0
1:4

Multiply both sides by 6

1 * 6 : 4 * 6

6:24

1:4 and 6:24 are proportional, so 1:4 and 6:18 are not.
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Write the equation for a line that has an initial value of 3 and 3/4 as it’s rate of change
frutty [35]

the slope goes by several names

• average rate of change

• rate of change

• deltaY over deltaX

• Δy over Δx

• rise over run

• gradient

• constant of proportionality

however, is the same cat wearing different costumes.

initial value of 3, namely when x = 0, y = 3, so we have the point (0 , 3) and it has a rate or slope of 3/4.

(\stackrel{x_1}{0}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{3}{4} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{3}{4}}(x-\stackrel{x_1}{0})\implies y=\cfrac{3}{4}x+3

3 0
3 years ago
Cindy set:<br> 2 groups of 10 buttons and 34 single buttons
vova2212 [387]
In all you have 54 buttons because if you add 10 plus 10 that equals 20 plus 34 and that gets 54 buttons in all
4 0
4 years ago
Calculate the slope of the line that passes through the points (-4,-7) and (1,-7)​
uranmaximum [27]

\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-7}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-7}-\stackrel{y1}{(-7)}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-4)}}}\implies \cfrac{-7+7}{1+4}\implies \cfrac{0}{5}\implies 0

8 0
4 years ago
Read 2 more answers
Can any kind soul help me ASAP!​
CaHeK987 [17]

Answer:

Hope it helps u

Step-by-step explanation:

As we know that ,

Mean = sum of the terms/ numbers of terms

But here grouped data is given so , we use the formula

Mean=∑[f. m]/ ∑f

where f is frequency and m is mid point of each height ,

Now first we have to find the mid point of each interval, where

midpoint of each interval = (lower boundary + upper boundary)/2

m1=(150+154)/2 = 152

m2=(155+159)/2= 157,now found other by same formula, for each interval

m3= 162

m4= 167

m5=172    Now we find the midpoint of each interval ,so now

∑[f. m]=f1*m1+f2*m2+f3*m3+f4*m4+f5*m5

now putting the values of each frequency for given interval and midpoint of each interval we will get,

∑[f. m]=456+942+1296+167*x+344 = 167*x+3038

Now find,

∑f=f1+f2+f3+f4+f5

∑f=19+x

Now we have,

∑[f. m]=167*x+3038

∑f=19+x

also given mean height=161.6 cm

putt these values in above equation we get,

161.6=\frac{167*x+3038}{19+x}

now solve this ,

161.6(19+x)=167*x+3038

3070.4+161.6*x=167*x+3038

3070.4-3038=167*x-161.6*x

32.4=5.4*x

x=32.4/5.4

<h2>x=6   Ans........</h2>

3 0
3 years ago
(4.7 * 108) + (3.35 x 10)
Bond [772]

Answer: 541.1

Step-by-step explanation:

multiply each set of parenthesis and add the two answers

7 0
3 years ago
Read 2 more answers
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