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3 years ago

What does the Roman numeral equal MCCCXLII

Mathematics

2 answers:

DedPeter [7]3 years ago

7 0

I got 1992...not sure though!‍♀️ I google searched it.

OLga [1]3 years ago

6 0

The answer would be 1342.

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Solve the equation for x, where x is a real number (5 points): <br> -11x^2 + 5x - 3 = 0

fenix001 [56]

To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
                   x = [ -b ± √(b^2 - 4ac) ] / (2a)
                   x = [ -5 ± √((5)^2 - 4(-11)(-3)) ] / ( 2(-11) )
                   x = [-5 ± √(25 - (132) ) ] / ( -22 )
                   x = [-5 ± √(-107) ] / ( -22)
Since we conclude that √-107 is nonreal, the answer to this question is that there are no real solutions.

7 0

3 years ago

Compute i^1+i^2+i^3....i^99+i^100

fgiga [73]

Good morning ☕️

Answer:

Step-by-step explanation:

Consider the sum S = i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰

S = i¹ + i² + i³ + . . . + i⁹⁹ + i¹⁰⁰

S = a₁ + a₂ + a₃ +. . . + a₉₉ + a₁₀₀

then, S is the sum of 100 consecutive terms of a geometric sequence (an)

where the first term a1 = i¹ = i and the common ratio = i

FORMULA:______________________

$S=(term1)*\frac{1-(common.ratio)^{number.of.terms}}{1-(common.ratio)}$

_______________________________

then

$S=i*\frac{1-i^{100} }{1-i}$

or i¹⁰⁰ = (i⁴)²⁵ = 1²⁵ = 1 (we know that i⁴ = 1)

Hence

S = 0

7 0

3 years ago

Find value of x 4(x+6)° (5x+11)°

Evgesh-ka [11]

Value of x = 13°

Hope it helps you...

5 0

3 years ago

Estimate the integral ∫6,0 x^2dx by the midpoint estimate, n = 6

Anettt [7]

Splitting up the interval [0, 6] into 6 subintervals means we have

$[0,1]\cup[1,2]\cup[2,3]\cup\cdots\cup[5,6]$

and the respective midpoints are $\dfrac12,\dfrac32,\dfrac52,\ldots,\dfrac{11}2$ . We can write these sequentially as ${x_i}^*=\dfrac{2i+1}2$ where $0\le i\le5$ .

So the integral is approximately

$\displaystyle\int_0^6x^2\,\mathrm dx\approx\sum_{i=0}^5({x_i}^*)^2\Delta x_i=\frac{6-0}6\sum_{i=0}^5({x_i}^*)^2=\sum_{i=0}^5\left(\frac{2i+1}2\right)^2$

Recall that

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$
$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$
$\displaystyle\sum_{i=1}^n1=n$

so our sum becomes

$\displaystyle\sum_{i=0}^5\left(\frac{2i+1}2\right)^2=\sum_{i=0}^5\left(i^2+i+\frac14\right)$
$=\displaystyle\frac{5(6)(11)}6+\frac{5(6)}2+\frac54=\frac{143}2$

8 0

3 years ago

Which zero pair could be added to the function so that the function can be written in vertex form?

Oduvanchick [21]

The function in this problem should be: <span>f(x) =x</span>² <span>+ 12x + 6

y = x</span>² + 12x + 6
y - 6 = x² + 12x

x² ⇒ x * x
12x ⇒ 2*6*x
missing number is 6² = 36

y - 6 + 36 = x² + 12x + 36

(x+6)(x+6) ⇒ x(x+6)+6(x+6) ⇒ x² + 6x + 6x + 36 = x² + 12x + 36

y + 30 = x² + 12x + 36
y = (x+6)² - 30

Choice is D. 36,-36

5 0

3 years ago