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3 years ago

Write the fraction in simplest from 8/10

Mathematics

2 answers:

AleksandrR [38]3 years ago

7 0

Divide common factors from both numerator and denominator.

Divide 2

8/2 = 4

10/2 = 5

8/10 simplified = 4/5

hope this helps

Sladkaya [172]3 years ago

5 0

Hi,

Answer:4/5

My work: This problem can be easily achieved by dividing the fraction 8/10 by 2.(8\10 / 2) When you do this you get the answer 4/5. To do this problem correctly you must divide not reduce and divide 8/2 and 10/2 this gets you the answer 4/5.

I Hope I Helped!

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Three circles with centers A, B, and C are externally tangent to each other as shown in the figure. Lines EG and DG are tangent

fiasKO [112]

My solution to the problem is as follows:

EC = 15 ... draw CF = 6 (radius) ...use Pythagorean theorem to find EF.

EF^2 + CF^2 = EC^2
EF^2 = 15^2 - 6^2 = 189 .... EF = sq root 189

triangle GDE is similar to CFE ... thus proportional
GD / ED = CF / EF
GD / 18 = 6 / (sq root 189)
GD = 108 / (sq root 189)

I hope my answer has come to your help. God bless and have a nice day ahead!

5 0

2 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$