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PSYCHO15rus [73]
3 years ago
12

Which is the solution to the equation 1/4x-1/8=7/8+1/2x

Mathematics
2 answers:
ivann1987 [24]3 years ago
5 0

Answer:

x = -4

Step-by-step explanation:

1/4x-1/8=7/8+1/2x

Add 1/8 to each side

1/4x-1/8+1/8=7/8+1/8+1/2x

1/4 x = 1 + 1/2x

Subtract 1/2x from each side

1/4x -1/2x = 1 + 1/2x -1/2x

Getting a common denominator for the x terms

1/4x -2/4x =1

-1/4x = 1

Multiply by -4

-4 * -1/4x = 1* -4

x = -4

timofeeve [1]3 years ago
4 0

Answer:

x = -4

Step-by-step explanation:

Step 1: Subtract 1/2x on both sides

-1/4x - 1/8 = 7/8

Step 2: Add 1/8 on both sides

-1/4x = 1

Step 3: Divide both sides by -1/4

x = -4

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Vedmedyk [2.9K]
Yeah what is the question
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3 years ago
Cynthia Besch wants to buy a rug for a room that is 21 ft wide and 34 ft long. She wants to leave
Daniel [21]

Answer:

Dimensions of the rug = 13 ft × 26 ft

Step-by-step explanation:

Dimensions of the room = 21 ft × 34 ft

Area of the room = 21 × 34 = 714 ft²

Cynthia wants to leave a uniform strip of floor around the rug.

Let the width of the rug = x ft

Then the dimensions of the rug will be = (21- 2x)ft × (34 - 2x)ft

Area of the rug = (21 - 2x)×(34 - 2x) square feet

338 = (21 - 2x)×(34 - 2x)

338 = 714 - 68x - 42x + 4x²

4x² - 110x + 714 - 338 = 0

4x² - 110x + 376 = 0

2x² - 55x + 188 = 0

2x² - 47x - 8x + 188 = 0

x(2x - 47) - 8(x - 47) = 0  

(x - 4)(2x - 47) = 0

x = 4, \frac{47}{2}

For x = 23.5 area of the rug will be negative.

Therefore, x = 4 ft will be the width of the rug.

Dimensions of the rug will be 13 ft × 26 ft. 

7 0
3 years ago
A stone is dropped from the top of a tower. The height h in meters from where the stone is dropped to the ground is given by the
CaHeK987 [17]

Wag umasa sa brainly

Step-by-step explanation:

Labot

5 0
2 years ago
Can someone please help me with this question?
sattari [20]
The answer would be y=1/7x-4
7 0
2 years ago
A motorboat takes 5 hours to travel 100mi going upstream. The return trip takes 2 hours going downstream. What is the rate of th
Temka [501]
The boat went 5hours upstream, let's say it has a "still water" speed rate of "b", it went 100miles... however, going upstream is going against the current, let's say the current has a speed rate of "c"

so, when the boat was going up, it wasn't really going "b" fast, it was going " b - c " fast, because the current was eroding speed from it

now, when coming down, the return trip, well the length is the same, so the distance is also 100miles, it only took 2hrs though, because, the boat wasn't coming down  "b" fast, it was coming down " b + c " fast, because the current was adding speed to it, so it came down quicker

now, recall your d = rt, distance = rate * time

\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
upstream&100&b-c&5\\
downstream&100&b+c&2
\end{array}
\\\\\\
\begin{cases}
100=(b-c)5\\
\qquad \frac{100}{5}=b-c\\
\qquad 20=b-c\\
\qquad 20+c=\boxed{b}\\
100=(b+c)2\\
\qquad 50=b+c\\
\qquad 50=\boxed{20+c}+c
\end{cases}

solve for "c", to see what's the current's speed

what's "b"?  well 20+c = b
4 0
3 years ago
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