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3 years ago

PLZ HELP PLZ! TY! WILL GIVE BRAINLIEST TO WHOEVER ANSWERS

Mathematics

1 answer:

Marysya12 [62]3 years ago

7 0

Take the y and x and 4 as a common factor and you will get 4xy(2yz+3x) the second one is the answer

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circle a has a radius of 7 and arcs rc cs and sm are congruent to the nearest hundereth what is the length of arc rc

oksian1 [2.3K]

The complete question in the attached figure

we know that
[arcs rc +arc cs arc sm]---------------> they take up half a circle-------->180°
 they are all congruent
so
divide 180 by 3 ----------------> 180/3=60°

Find the length of arc rc=60°

if 360°----------------> (2*pi*7) length of circumference
60° -------------------> X
X=60*2*pi*7/360------------> 7.327 units

the answer is
the length of arc rc= 7.33 units

3 0

3 years ago

121 over 121 all the names that apply to each number

spayn [35]

121 over 121 is a whole number of 1

6 0

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