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4 years ago

What is the product of the two factors? (X 4/3) (X 2/3)

Mathematics

2 answers:

SpyIntel [72]4 years ago

5 0

Answer:

$\frac{8}{9}{x}^{2}$

Step-by-step explanation:

All you do is multiply straight across [both denominator and numerator] to arrive at your answer. Then, multiplying two x's together gives you ${x}^{2}$ . So, with all that being said, you have your answer.

I am joyous to assist you anytime.

DochEvi [55]4 years ago

3 0

Answer:x^2

Step-by-step explanation:

I did what the comment said and it was wrong but I watched the video after I got it wrong on algebra nation and it said x^2 was the answer

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Help me with this problem please right away

svlad2 [7]

Answer:

$6\frac{1}{2} = A$

Step-by-step explanation:

The first step is to use the Distance Formula to figure out the lengths of the two legs:

$\sqrt{[-x_1 + x_2]^{2} + [-y_1 + y_2]^{2}} = D$

R(1, 1) and P(−1, −2) ↷

$\sqrt{[1 + 1]^{2} + [2 + 1]^{2}} = \sqrt{2^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13}\\ \\ \sqrt{13} = RP$

Q(2, −4) and P(−1, −2) ↷

$\sqrt{[1 + 2]^{2} + [2 - 4]^{2}} = \sqrt{[-2]^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13} \\ \\ \sqrt{13} = QP$

* So, by my calculations, we have an isosceles right triangle, and according to the 45°-45°-90° triangle theorem, we automatically know that the hypotenuse, RQ, is $\sqrt{26}$ :

30°-60°-90° triangle theorem

2x, x, x√3

↑ ↑ ↑

h leg leg

45°-45°-90° triangle theorem

x√2, x, x

↑ ↑ ↑

h legs

So now, we have to find the area of the triangle by taking half of either height or base, then multiplying that by the height or base, but since this is an isosceles right triangle, it does not matter:

$[\frac{h}{2}][b] = A \: \:OR\: \: [\frac{b}{2}][h] = A \: \:OR \: \: \frac{hb}{2} = A$

$[\frac{\sqrt{13}}{2}][\sqrt{13}] = \frac{13}{2} = 6\frac{1}{2} \\ \\ 6\frac{1}{2} = A$

I am joyous to assist you anytime.

8 0

3 years ago

¿Qué significa un punto en el diagrama de líneas?

Nostrana [21]

un punto en un gráfico de líneas es un valor específico que normalmente es un valor de entrada y salida (x)

7 0

3 years ago

You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you want to place it whe

strojnjashka [21]

Answer:

Step-by-step explanation:

$\text{The equation of the surface of the sphere whose radius is 6 can be represented as:}$

$x^2+y^2+z^2 = 36}$

$So , g(x,y,z) = x^2 + y^2 + z^2 - 36$

$\text{To minimize the function :} \\ \\ M(x,y,z0 = 6x - y^2 +xz + 60$

$\text{by applying lagrange multipliers }$

$\bigtriangledown M (x,y,z) = \lambda \bigtriangledown g(x,y,z)$

$\langle 6+zz,-2y,x \rangle = \lambda \langle 2x,2y,2z \rangle$

$6+z = 2\lambda x$

$-2y = 2 \lambda y$

$\text{from the second equation ;} (\lambda +1)y = 0, \text{so , it is either y =0 or }\lambda = -1$

$Suppose \ \lambda = -1 ; \text{other equation becomes x = -2z and 6+z = -2x}$

$\text{such that; 6+z =4z or z = 2}$

$\text{it implies that: x= -4 and }y = \pm \sqrt{36 - 4-16} = \pm 4$

$\text{Here; there exist two possible points}$

$\text{(-4,-4,2) and (-4,4,2)}$

$\text{In the scenario here y=0,} \lambda \text{ is unknown, then we remove it}$

$x = 2 \lambda z \\ \\ 6+z = 2 \lambda x \\ \\ 6+z = 4 \lambda ^2z \\ \\ 6 = (4 \lambda ^2 -1) z ---(1)$

$z = \dfrac{6}{4 \lambda ^2 -1}$

$x = \dfrac{12 \lambda }{4 \lambda ^2 -1 }$

$\text{recall that; it i possible to divide }4 \lambda ^2 -1 \text{since (1) shows that it cannot be equal to zero.} \\ \\ \text{hence, puttinf this into constraint, whereby y =0, Then:}$