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3 years ago

Can someone help me on this math

Mathematics

1 answer:

Tomtit [17]3 years ago

5 0

58.96 or 1887/32 id the answer

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Which of the following equations represents the line with a slope of -7/4 and a y-intercept of -11?

andreev551 [17]

Y= -7/4x - 11
My answer has to be 20 characters longkdmmwmmckkdmsnskckd

4 0

3 years ago

What is the derivative of x times squaareo rot of x+ 6?

Dafna1 [17]

Hey there, hope I can help!

$\mathrm{Apply\:the\:Product\:Rule}: \left(f\cdot g\right)^'=f^'\cdot g+f\cdot g^'$
$f=x,\:g=\sqrt{x+6} \ \textgreater \ \frac{d}{dx}\left(x\right)\sqrt{x+6}+\frac{d}{dx}\left(\sqrt{x+6}\right)x \ \textgreater \ \frac{d}{dx}\left(x\right) \ \textgreater \ 1$

$\frac{d}{dx}\left(\sqrt{x+6}\right) \ \textgreater \ \mathrm{Apply\:the\:chain\:rule}: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \ \textgreater \ =\sqrt{u},\:\:u=x+6$
$\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+6\right)$

$\frac{d}{du}\left(\sqrt{u}\right) \ \textgreater \ \mathrm{Apply\:radical\:rule}: \sqrt{a}=a^{\frac{1}{2}} \ \textgreater \ \frac{d}{du}\left(u^{\frac{1}{2}}\right)$
$\mathrm{Apply\:the\:Power\:Rule}: \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \ \textgreater \ \frac{1}{2}u^{\frac{1}{2}-1} \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{u}}$

$\frac{d}{dx}\left(x+6\right) \ \textgreater \ \mathrm{Apply\:the\:Sum/Difference\:Rule}: \left(f\pm g\right)^'=f^'\pm g^'$
$\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(6\right)$

$\frac{d}{dx}\left(x\right) \ \textgreater \ 1$
$\frac{d}{dx}\left(6\right) \ \textgreater \ 0$

$\frac{1}{2\sqrt{u}}\cdot \:1 \ \textgreater \ \mathrm{Substitute\:back}\:u=x+6 \ \textgreater \ \frac{1}{2\sqrt{x+6}}\cdot \:1 \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{x+6}}$

$1\cdot \sqrt{x+6}+\frac{1}{2\sqrt{x+6}}x \ \textgreater \ Simplify$

$1\cdot \sqrt{x+6} \ \textgreater \ \sqrt{x+6}$
$\frac{1}{2\sqrt{x+6}}x \ \textgreater \ \frac{x}{2\sqrt{x+6}}$
$\sqrt{x+6}+\frac{x}{2\sqrt{x+6}}$

$\mathrm{Convert\:element\:to\:fraction}: \sqrt{x+6}=\frac{\sqrt{x+6}}{1} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}}{1}$

Find the LCD
$2\sqrt{x+6} \ \textgreater \ \mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}\cdot \:2\sqrt{x+6}}{2\sqrt{x+6}}$

$Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions$
$\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{x+2\sqrt{x+6}\sqrt{x+6}}{2\sqrt{x+6}}$

$x+2\sqrt{x+6}\sqrt{x+6} \ \textgreater \ \mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c}$
$\sqrt{x+6}\sqrt{x+6}=\:\left(x+6\right)^{\frac{1}{2}+\frac{1}{2}}=\:\left(x+6\right)^1=\:x+6 \ \textgreater \ x+2\left(x+6\right)$
$\frac{x+2\left(x+6\right)}{2\sqrt{x+6}}$

$x+2\left(x+6\right) \ \textgreater \ 2\left(x+6\right) \ \textgreater \ 2\cdot \:x+2\cdot \:6 \ \textgreater \ 2x+12 \ \textgreater \ x+2x+12$
$3x+12$

Therefore the derivative of the given equation is
$\frac{3x+12}{2\sqrt{x+6}}$

Hope this helps!

8 0

3 years ago

Assuming that one-pint is equal to 473 ml, how many pints can be found in 3 liters?

Svetlanka [38]

Given that 1 pint is equal to 437 ml
and 1000 ml=1 liter
then
1 pint =(437/1000) liters
simplifying we get:
1 pint= 0.437 liters
thus amount of pints in 4 liters will be:
4/0.473
=8.45666586
~8.5 pints

4 0

3 years ago

A regular hexagon with side length 4 has the same area as a square. What is the length of the square?

svetlana [45]

Answer: The length of the square is $6.44$

Step-by-step explanation:

By definition, the area of a Regular polygon can be calculated with the following formula:

$A=\frac{s^2n}{4tan(\frac{180}{n})}$

Where "s" the length of any side of the polygon and "n" is the number of sides .

According to the information given in the exercise, you know that:

$s=4$

Since an hexagon has six sides, you know that:

$n=6$

Therefore, its area is:

$A_h=\frac{(4^2)(6)}{4tan(\frac{180}{6})}\\\\A_h=24\sqrt{3}$

The formula to find the area of a square is:

$A=s^2$

Where "s" is the length of any side of the square.

Since that regular hexagon has the same area as this square, you can substituting the area calculated above into the formula for calculate the area of a square, and then solve for "s".

Then you get:

$24\sqrt{3}=s^2\\\\\sqrt{24\sqrt{3}}=s\\\\s=6.44$

7 0

3 years ago

I need help with #6!! Thanks

tatyana61 [14]

Use this formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

a) d=11

Work:
d=sqrt((4-(-7))^2+(5-5)^2
d=sqrt((11)^2+(0)^2)
d=sqrt(121+0)
d=sqrt(121)
d=11

b) d=4

c) d = 16

d) d=160

e) d=12

6 0

3 years ago