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bazaltina [42]
3 years ago
6

Write 3 1/8 as a decimal.

Mathematics
2 answers:
Alenkinab [10]3 years ago
5 0
No it is not. It is a fraction
VladimirAG [237]3 years ago
3 0
The answer is 3.125
That's correct...
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Solve using perfect square factoring patterns. x2 – 12x + 36 = 4 A. {–8, –4} B. {–2, 10} C. {4, 8} D. {10, –2}
Nataliya [291]
<span> x2 – 12x + 36 = 4
(x - 6)^2 = 2^2

x - 6 = 2
x = 8

-(x - 6) = 2
-x + 6 = 2
-x = -4
x = 4

answer
</span><span>C. {4, 8}</span>
3 0
3 years ago
How do I calculate this? Is there a formula?
alex41 [277]

<u>Answer:</u>

Height of cables = 23.75 meters

<u>Step-by-step explanation:</u>

We are given that the road is suspended from twin towers whose cables are parabolic in shape.

For this situation, imagine a graph where the x-axis represent the road surface and the point (0,0) represents the point that is on the road surface midway between the two towers.

Then draw a parabola having vertex at (0,0) and curving upwards on either side of the vertex at a distance of x = 600 or x = -600, and y at 95.

We know that the equation of a parabola is in the form y=ax^2 and here it passes through the point (600, 95).

y=ax^2

95=a \times 600^2

a=\frac{95}{360000}

a=\frac{19}{72000}

So new equation for parabola would be y=\frac{19x^2}{72000}.

Now we have to find the height (y)of the cable when x= 300.

y=\frac{19 (300)^2}{72000}

y = 23.75 meters

8 0
3 years ago
Read 2 more answers
Solve for x: 5/8=x-1/9<br><br> A. 37/8<br> B. 23/4<br> C. 11/2<br> D. 53/8
Vesna [10]

Answer:

53/72

Step-by-step explanation:

5/8 = x - 1/9

+1/9       +1/9

5/8 + 1/9 = x

9(5/8)  + 8(1/9)

45/72 + 8/72

x= 53/72

8 0
2 years ago
If Rupert does all his math quizzes, and if he scores 4 correct for every 5 questions, what will his percentage correct be?
Sauron [17]

Answer:

LOL IT'S GONNA BE 80%

Step-by-step explanation:

HOPE THIS HELPS, SORRY FOR ALL THE CAPS

5 0
2 years ago
6) Sally wrote the number 30, 048 in expanded form.
vichka [17]

Answer:

the answer is G) Change 30 to 3,000

6 0
3 years ago
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