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Inessa [10]
3 years ago
7

Math help please will mark brainliest

Mathematics
1 answer:
balu736 [363]3 years ago
4 0

Answer:

you got there answer correct

Step-by-step explanation:

the lines all follow the answer

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Which of the following expressions could be used to represent "the difference between -17 and -8"?
musickatia [10]
The  last bc the + and - are different that means it will be subtracted.
7 0
3 years ago
Exactly 1 1/3 yard of ribbon is needed to make a bow. Which of the following lengths of ribbon could be used to make a bow with
svp [43]

Answer:

1 and 2/5 yards wastes only 1/15 a yard

Step-by-step explanation:

1 and 1/6 yards  would make a waste of 1 and 1/6 yards since you can't make a ribbon out of it. (incorrect)

1 and 2/10 yards  would make a waste of 1 and 1/5 yards since you can't make a ribbon out of it. (incorrect)

1 and 2/5 yards

1\frac{2}{5} -1\frac{1}{3} =\\\\1\frac{6}{15} -1\frac{5}{15} =\\\\\frac{1}{15}

Wastes 1/15 a yard.

1 and 1/2 yards  

1\frac{1}{2} -1\frac{1}{3} =\\\\1\frac{3}{6} -1\frac{2}{6} =\\\\\frac{1}{6}

It wastes 1/6 a yard.

2 yards

2-1\frac{1}{3} =\frac{2}{3}

It wastes 4/6 a yard.

4 0
3 years ago
Help me, please? I don't get it
Korvikt [17]

Answer:

1/10 = 2/20

1/3 = 3/9

5/7 = 15/21

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
the same type of television is being sold at two different stores. store A is selling the television for $800, which is $120 mor
svlad2 [7]

Answer: the selling price of the television at store B is $1360

Step-by-step explanation:

Assuming the selling price of the television at store B is represented by $B. Store A is selling the television for $800, which is $120 more than half the cost of the television at store B. The equation representing this situation is expressed as

800 = B/2 + 120

Multiplying the left hand side and the right hand side of the equation by 2, it becomes

1600 = B + 240

B = 1600 - 240

B = $1360

5 0
3 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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