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3 years ago

Need help:(.... Simplify (b^5)^2

Mathematics

2 answers:

olga55 [171]3 years ago

7 0

For this case we must indicate an expression equivalent to:

$(b ^ 5) ^ 2$

We have that by definition of properties of powers that is fulfilled:

$(a ^ n) ^ m = a ^ {n * m}$

Now, we can rewrite the given expression as:

$(b ^ 5) ^ 2 = b ^ {5 * 2} = b ^ {10}$

ANswer:

$b^{10}$

Option A

Charra [1.4K]3 years ago

5 0

b^10 would be the correct answer :)

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-t+5=t-19<br> Need an answer and a work pls!

Nadusha1986 [10]

Answer:

t=12

Step-by-step explanation:

Step 1: Subtract t from both sides.

−t+5−t=t−19−t

−2t+5=−19

Step 2: Subtract 5 from both sides.

−2t+5−5=−19−5

−2t=−24

Step 3: Divide both sides by -2.

−2t−2=−24−2

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3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

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4 years ago

A block of concrete is 1m long, 50cm wild and 4cm thick. 1.calculate the volume of the block in cm3 .2.if 1cm has a mass of 2.7g

Nonamiya [84]

9514 1404 393

Answer:

20,000 cm³
54 kg

Step-by-step explanation:

1 m is 100 cm, so the volume is ...

V = LWH

V = (100 cm)(50 cm)(4 cm) = 20,000 cm³ . . . volume

The mass is ...

(20,000 cm³)(2.7 g/cm³) = 54,000 g = 54 kg

The mass of the block is 54 kg.

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11Alexandr11 [23.1K]

The answer to your question is D B and G.

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Darina [25.2K]

Answer:

False

Step-by-step explanation:

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3 years ago