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borishaifa [10]
3 years ago
8

Look at pic will mark brainilest

Mathematics
2 answers:
Aliun [14]3 years ago
6 0
The answer is the second one 40inches
densk [106]3 years ago
5 0

Answer:The first one 18in.2

Step-by-step explanation:

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Find the measure of angle 1.
Vesna [10]

Answer:

awesome

Step-by-step explanation:

3 0
3 years ago
When it comes to reducing that’s where I start having trouble for example 10/13x1/15= 10/65 but 10/65 not in lowest terms so I h
arlik [135]

Answer:

Step-by-step explanation:

Let's use your example as a starting point.  <em>Determine whether the same number will divide each 10 and 65 evenly</em>.  In this case, the answer is yes, and the number is 5.  10/5 = 2, and 65/5 = 13.  Thus, the fraction in lowest terms is 2/13.

6 0
3 years ago
I will do anything plss asppp
AlexFokin [52]

Answer:

I think the answer is A

Step-by-step explanation:

Let me know if it is not

4 0
3 years ago
What would you have to do to change 10 cubic feet into cubic inches? 
Temka [501]
1 ft = 12 inches
(1 ft)^3 = (12 inches)^3
1 cubic foot = 1728 cubic inches

You'll multiply by the conversion factor (1278 in^3)/(1 ft^3) to convert from cubic feet to cubic inches. Notice how the cubic feet ft^3 unit is at the bottom of the conversion fraction, so that the cubic feet units cancel.

Answer: A) multiply by 1728
3 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
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