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3 years ago

Which of the following equations is equivalent to S = pi r squared h?

Mathematics

1 answer:

skad [1K]3 years ago

8 0

Answer:

Step-by-step explanation:

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I don’t understand this either

trasher [3.6K]

It is a part of slope form ( intercept

7 0

3 years ago

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Find the solution to the system of equations: x + 3y = 7 and 2x + 4y = 8 1. Isolate x in the first equation: 2. Substitute the v

klemol [59]

X + 3y = 7
x = -3y + 7

2x + 4y = 8
2(-3y + 7) + 4y = 8
-6y + 14 + 4y = 8
-2y = 8 - 14
-2y = - 6
y = -6/-2
y = 3

x + 3y = 7
x + 3(3) = 7
x + 9 = 7
x = 7 - 9
x = -2

solution is (-2,3)

6 0

3 years ago

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What is the value of the expression 6+18 % 3x4

natka813 [3]

You tell me buddy :)

3 0

3 years ago

Which of the following is equivalent to 2 ln e^ln 5x = 2 ln 15?

neonofarm [45]

Answer:

x = 3

Step-by-step explanation:

Given in the question an equation,

$2lne^{ln5x}=2ln15$

Step 1

$e^{lnx}=x$

so,

$2ln(5x)=2ln15$

Step 2

cancel 2 on both sides of the equation

$ln{5x}=ln15$

Step 3

$ln{5x}-ln15=0$

$ln\frac{5x}{15}=0$

Step 4

$ln\frac{x}{3}=0$

Step 5

$e^{ln\frac{x}{3}}=e^{0}$

Step 6

x/3 = 1

x = 3(1)

x = 3

8 0

2 years ago

Given that −4i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. f

GalinKa [24]

Answer:

$\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }$

Step-by-step explanation:

Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).

$\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}$

The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.

$f(x)=x^4-2x^3+x^2-32x-240\\\\ =(x^2+16)(x^2+ax-15)\\\\ =x^4+ax^3-15x^2+16x^2+16ax-240$

We identify the coefficients for the like terms, it comes

a = -2 and 16a = -32 (which is equivalent). So, we can write in $\mathbb{R}$ .

$\\f(x)=(x^2+16)(x^2-2x-15)$

The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.

$f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}$

And we can write in $\mathbb{C}$

$f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

7 0

2 years ago