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Lynna [10]
3 years ago
7

What is the value of x in 6-3x=5x-10x+12

Mathematics
1 answer:
Alexus [3.1K]3 years ago
5 0
3 is the right answer
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Find the angle measure C in the parallelogram ABCD.​
Kryger [21]

Answer:

135°

Step-by-step explanation:

In a parallelogram the opposite angles are congruent, so

∠BCD = ∠BAD = 135°

5 0
3 years ago
I need to subtract it with renaming it too.
Vladimir79 [104]
8 \frac{1}{6} -3 \frac{5}{6} = 7 \frac{7}{6} -3 \frac{5}{6}
= 4 \frac{2}{6} = 4 \frac{1}{3}
6 0
3 years ago
Read 2 more answers
ZX is perpendicular bisector of WY if WY=24 what is WX
bixtya [17]

Answer:

12

Step-by-step explanation:

6 0
3 years ago
In a cave, a stalactite gets 44 millimeters longer each year. this year it is 72 centimeters long. how many years until it is 1
iragen [17]

Answer:

6.36 years (the .36 repeats), so rounded is 6.4.

Step-by-step explanation:

First, I'm going to convert all measurements to meters.

Since 1000 mm = 1 m, divide 44 by 1000.

44/1000 = 0.044 m

Since 100 cm = 1 m, divide 72 by 100.

72/100 = 0.72 m

Next, I'm going to setup an equation using y = mx + b.

b is the beginning amount which is 0.72.

m in the equation represents the rate which is 0.044.

y represents the total which is 1.

Plugin the numbers and solve for x.

1 = 0.044x + 0.72

1 - 0.72 = 0.044x + 0.72 - 0.72

0.28 = 0.044x

0.28/0.044 = 0.044x/0.044

6.36 = x

7 0
2 years ago
Petes boat can travel 48 miles upstream in 4 hours. The return trip takes 3 hours. Find the speed of the boat in still water and
tankabanditka [31]
So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up

going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r

notice, the distance is the same, upstream as well as downstream
thus   \bf \begin{cases}
b=\textit{rate of the boat}\\
r=\textit{rate of the river}
\end{cases}\qquad thus
\\\\\\

\begin{array}{lccclll}
&distance&rate&time(hrs)\\
&----&----&----\\
upstream&48&b-r&4\\
downstream&48&b+4&3
\end{array}
\\\\\\

\begin{cases}
48=(b-r)(4)\to 48=4b-4r\\\\
\frac{48-4b}{-4}=r\\
--------------\\
48=(b+r)(3)\\
-----------------------------\\\\
thus\\\\
48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3)
\end{cases}

solve for "r", to see what the stream's rate is

what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
5 0
3 years ago
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