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3 years ago

Choose the product. a 3 b 2 · 4ab 3 4a3b5 4a3b6 4a4b5

1 answer:

5 0

A³ b² 4ab³

Rearrange order:

4 a³ a b² b³

Now add up the exponents from same base:

4 a³⁺¹ b²⁺³

4 a⁴ b⁵

Final answer: 4 a⁴ b⁵

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What’s the volume of a cylinder with a radius of 6ft and height of 14ft? pls show work

Daniel [21]

Answer:

Given -

radius of cylinder = 6 ft
height of cylinder = 14 ft

Now ,

$volume = \pi \: r {}^{2} h \\ = > \frac{22}{\cancel{7}} \times 6 \times\cancel{ 14} \\ \\ = > 22 \times 6 \times 2 \\ = > 264 \: ft {}^{3}$

hope helpful~

5 0

2 years ago

A cable pole casts a shadow that is 12 feet long . If a 6ft tall man casts a shadow thats is 4 feet long how tall is the cable p

shepuryov [24]

Answer:

18ft

Step-by-step explanation:

6ft -> 4ft shadow

6 = 4m

m = 6/4

y = 12m

y = 12(6/4)

y = 18ft

4 0

3 years ago

If cos θ > 0, which quadrant(s) could the terminal side of θ lie?

Sergeeva-Olga [200]

Answer:

I and IV

Step-by-step explanation:

cosine is positive/greater than 0 in quadrants I and IV. Its just a rule of the unit circle.

Can remember by All Star Trig Class

meaning all the trig funtions are positive in quadrant I

sin is positive in II

tan is positive in III

and cos is positive in IV

8 0

3 years ago

Find the sum 2.75+(-2)+(5.25)

viva [34]

Answer:

6 is the answer

Step-by-step explanation:

Hope I have helped.

8 0

2 years ago

Read 2 more answers

Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2

Finger [1]

Answer:

$\frac{y}{x^2}=\sin x+\pi$

Step-by-step explanation:

Consider linear differential equation $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$

It's solution is of form $y\,I.F=\int I.F\,q(x)\,dx$ where I.F is integrating factor given by $I.F=e^{\int p(x)\,dx}$ .

Given: $\frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x$

We can write this equation as $\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x$

On comparing this equation with $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$ , we get $p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x$

I.F = $e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}$ { formula used: $\ln a^b=b\ln a$ }

we get solution as follows:

$\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C$

{ formula used: $\int \cos x\,dx=\sin x$ }

Applying condition: $y(\pi)=\pi^2$

$\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C$

So, we get solution as :

$\frac{y}{x^2}=\sin x+\pi$

4 0

3 years ago