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2 years ago

What are the like fractions of 1/2 and 3/7

Mathematics

1 answer:

Step2247 [10]2 years ago

8 0

Answer:

7/14 and 6/14

Step-by-step explanation:

we are trying to find the same denominator for the 2 fractons, so 1/2 times 2 to both sides would be 7/14. and 3/7 times 2 to both sides would be 6/14. (same denominator)

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What is the average of 63 and 148

Veseljchak [2.6K]

The average is 105.5

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3 years ago

Bianca walks her dog everyday. She walks 0.6 mile on each weekday. On each weekend day, the walk is 1 1/2 times as long as a wal

mixas84 [53]

Answer: 4.8

Step-by-step explanation:0.6*5+0.6*1 1/2*2

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3 years ago

Divide the number 720 into the following Ratios<br><br>2:1:3<br><br>5:4

solmaris [256]

Answer:

Ratio 1 = a(240

=b)120

=c)360

Ratio 2=a)400

=b)320

Step-by-step explanation:

On both of the ratios you first add them then divide one by one like this for example 2+1+3= 6

then 2/6 ÷720 = 120 ×2 =240 then you do this to the other numbers as well.

6 0

2 years ago

Read 2 more answers

Solve the equation 14z=224

kolbaska11 [484]

Answer:

Step-by-step explanation:

224/14=16

6 0

3 years ago

Read 2 more answers

Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Spr

NeX [460]

Answer:

20.2057 Units.

Step-by-step explanation:

First, we determine the length of the cable.

Distance between Centerville (8,0) and point (x,0) is given as:

$\sqrt{(8-x)}^2=8-x$

Distance between point (x,0) and Springfield(0,7) is:

$\sqrt{(7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Distance between point (x,0) and Shelvyfield(0,-7) is:

$\sqrt{(-7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Therefore the Length of the Cable L(x)

$L(x)=(8-x)+2\sqrt{(7)^2+(x)^{2}}$

To find the critical point, we set the derivative of L(x)=0

$L^{'}(x)=-1+\frac{2x}{\sqrt{\left( 49 - x^{2}\right) }}$

$\frac{-\sqrt{\left( 49 - x^{2}\right)}+2x}{\sqrt{\left( 49 - x^{2}\right)}}=0\\-\sqrt{\left( 49 - x^{2}\right)}+2x=0\\\sqrt{\left( 49 - x^{2}\right)}=2x\\(\sqrt{\left( 49 - x^{2}\right)})^2=(2x)^2\\49 - x^{2}=4x^2\\49=5x^2\\x^2=\frac{49}{5}\\x= 3.1305$

To verify that L(x) has a minimum at this critical number we compute the second derivative L″(x) and find its value at the critical number.

$L^{''}(x)=\frac{\left( 98\right) \,\sqrt{\left( 49 - x^{2}\right) }}{{\left( 7 - x\right) }^{2}\,{\left( 7+x\right) }^{2}}\\At \:x=3.1305, L^{''}=0.3993$

Since $L^{''}(x)$ is positive, the minimum point of L(x) exists.

Next, we find the minimum length by substituting z=3.1305 into L(x)

$L(3.1305)=(8-3.1305)+2\sqrt{(7)^2+(3.1305)^{2}}$

Minimum Length, L=20.2057

The minimum length of the cable is 20.2057 Units.

6 0

3 years ago