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3 years ago

Please, its for a test!

Mathematics

1 answer:

-BARSIC- [3]3 years ago

8 0

Answer:

Step-by-step explanation:

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what is the formula for to find the surface area of a triangular prism and which one is the height and which one is the zbase

Nina [5.8K]

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4 years ago

672,591 is closer to 700,000 than

GaryK [48]

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3 years ago

Read 2 more answers

What is the product? Two-thirds times eight-ninths StartFraction 10 over 27 EndFraction StartFraction 16 over 27 EndFraction Two

melomori [17]

Answer:

16/27

Step-by-step explanation:

First let's convert this info an equation and then we will solve.

(2/3) * (8/9) = ?

( 2 * 8 ) / ( 3 * 9 ) = ?

( 16 ) / ( 27 ) = ?

16 / 27 = ?

So our fractional answer is 16/27.

In words this would be sixteen over twenty seven.

Cheers.

3 0

3 years ago

What is the principal square root of -4

devlian [24]

Answer:

The principal square root of -4 is 2i.

Step-by-step explanation:

$\sqrt{-4}$ = 2i

We have the following steps to get the answer:

Applying radical rule $\sqrt{-a} =\sqrt{-1} \sqrt{a}$

We get $\sqrt{-4} =\sqrt{-1} \sqrt{4}$

As per imaginary rule we know that $\sqrt{-1}=i$

= $\sqrt{4} i$

Now $\sqrt{4} =2$

Hence, the answer is 2i.

6 0

3 years ago

Solve the following ODE's: c) y* - 9y' + 18y = t^2

Nastasia [14]

Answer:

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

Step-by-step explanation:

y''- 9 y' + 18 y = t²

solution of ordinary differential equation

using characteristics equation

m² - 9 m + 18 = 0

m² - 3 m - 6 m+ 18 = 0

(m-3)(m-6) = 0

m = 3,6

C.F. = $C_1e^{3t}+C_2e^{6t}$

now calculating P.I.

$P.I. = \frac{t^2}{D^2 - 9D +18}$

$P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

hence the complete solution

y = C.F. + P.I.

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

7 0

3 years ago