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BabaBlast [244]
3 years ago
11

What is 23,556 to the nearest ten thousand

Mathematics
2 answers:
maxonik [38]3 years ago
5 0
The nearest to the ten thousand is 20,000.
hodyreva [135]3 years ago
5 0

Answer:

20,000

Step-by-step explanation:

because 3 is less than 5 so you round down

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The answer is ten million six hundred thousand.
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If x=7 and y=3 what is the value of x+5y<br><br> i need help
aksik [14]

If x=7 and y=3, then

x+5y=7+5(3)

5(3)=15

15+7=22

So x+5y=22

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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
2 years ago
Pls if anyone knows the answer with work included/steps that will be greatly appreciated :)
dimaraw [331]

Answer:

1. Option D. 15x²

2. Option C. 3

Step-by-step explanation:

1. Determination of the area of one section.

Length (L) of one section = 25x/5 = 5x

Width (W) of one section = 3x

Area (A) of one section =?

The area of one section can be obtained as follow:

Area (A) = Length (L) × Width (W)

A = L × W

A = 5x * 3x

A = 15x²

Thus, the area of one section is 15x²

2. Determination of the expressions that are equivalent to (p²)³.

We'll begin by simplifying (p²)³. This can be obtained as follow:

(p²)³ = p²*³

(p²)³ = p⁶

Next we shall compare each expression given in the question above to see which will be the same as p⁶.

p × p × p × p × p × p = p¹⁺¹⁺¹⁺¹⁺¹⁺¹

p × p × p × p × p × p = p⁶

p² × p² × p² = p²⁺²⁺²

p² × p² × p² = p⁶

p² × p³ = p²⁺³

p² × p³ = p⁵

Thus,

p² × p³ ≠ p⁶

p⁵ ≠ p⁶

p⁶ = p⁶

SUMMARY

p × p × p × p × p × p = p⁶ = (p²)³

p² × p² × p² = p⁶ = (p²)³

p⁶ = (p²)³

Therefore, 3 expressions are equivalent to (p²)³. Option C gives the correct answer to the question.

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