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julia-pushkina [17]
2 years ago
9

4 + 3 + 7 to the 3rd power

Mathematics
2 answers:
bagirrra123 [75]2 years ago
8 0
4+3=7 7+7=14 14*14=196 196*14=2744, sooo it's 2744

lesya [120]2 years ago
4 0
The answer is 2744 because 4+3+7=14then 14x14x14=2744
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43 ones x 3 ten equals
Svet_ta [14]
The answer is 1230!!!!!!!
4 0
3 years ago
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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
3 years ago
Verify the identity by transforming the left-hand side into the right-hand side.
qaws [65]
<span>(1 + cos² 3θ) / (sin² 3θ) = 2 csc² 3θ - 1

Starting with the left:  Note that  cos²θ + </span><span>sin²θ = 1.
In the same way:    </span><span>cos²3θ + <span>sin²3θ = 1
</span></span>Therefore                cos²3θ  = 1 - <span>sin²3θ
</span> From the top:  (1 + cos² 3θ) = 1 + 1 - sin²3θ = 2 - <span>sin²3θ
</span>
(1 + cos² 3θ) / (sin² 3θ) = (<span>2 - sin²3θ) / (sin² 3θ) = 2/</span><span>sin² 3θ - </span><span>sin²3θ/</span>sin²3θ

=  2/<span>sin² 3θ - 1;        But 1/</span><span>sinθ = csc</span><span>θ,  Similarly  </span>1/sin3θ = csc3θ

= 2 *(1/sin<span>3θ)²  - 1</span>
= 2csc²3θ  - 1.        Therefore  LHS = RHS.  QED.

8 0
3 years ago
Solve: −4d − 6 = −10
Stels [109]

Answer:

hi

Step-by-step explanation:

- 4d - 6 =  - 10 \\  - 4d =  - 10 + 6 \\  - 4d =  - 4 \\ d = 1

i think so…

8 0
3 years ago
Read 2 more answers
Solve -17.6 - (-21.3)
AleksAgata [21]
-17.6 -(-21.3) = -17.6 + 21.3=3.7
3 0
3 years ago
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