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Artyom0805 [142]
3 years ago
12

Pepe uses these steps to partition segment AB in a ratio of 2:5.

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
7 0

Answer:

Step-by-step explanation:

the 1st one

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Can u pls help me with this question asap
Zinaida [17]

Answer:

3.5/1 (or just 3.5)

Step-by-step explanation:

If y = a/n, then the constant proportionality is 3.5/1 or 7/2

Think of this as "rise over run" (we are basically finding the slope)

7 0
2 years ago
Read 2 more answers
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
Help with the linear equations
IgorC [24]

Answer:

y \: intercept =  - 1 \frac{1}{2}

Step-by-step explanation:

The y intercept form for the equation of a line is

y = mx + c

You should note that <em>c</em> represents the y-intercept of the line (where the line touches the y-axis)

y = mx + c \\ \\ we \:were\: given \:the \:equation \:y = x - \frac{3}{2} \\ \\ therefore \: the\: value \:of \:c \:is\: - \frac{3}{2} \:or -1 \frac {1}{2}

7 0
2 years ago
I need these all written out on how they're done along with the answer please
taurus [48]
Well, these are quite simple really, especially 1,2,3 and 4. If you read the questions carefully, you can understand it well. You don’t need to know the answer, you just need to understand
8 0
3 years ago
Given the points A(-1, 2) and B(7,6), find the coordinates of the point P on the
Sergio039 [100]

Answer:

plot ab is right just had the test

6 0
3 years ago
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