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3 years ago

X^2=20 square root property

1 answer:

4 0

${x}^{2} = 20 \\ x = \sqrt{20} \\ x = \sqrt{4} \times \sqrt{5} = 2 \times \sqrt{5} \\ \\ x = 2 \sqrt{5} \\ therefore \: \: x = 2\sqrt{5}$

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Answer:

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3 years ago

Assume you are rolling two dice; the first one is red, and the second one is green. Use a systematic listing to determine the

Ede4ka [16]

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4 years ago

Surface area of solid cuboid 8cm 6cm 2cm

strojnjashka [21]

Answer:

152 $\text{cm}^2$

Step-by-step explanation:

We know that the total surface area of a cuboid is the sum of the surface area of all six of its faces.

We also know that each opposing side has the same surface area. Thus the total surface area is equal to:

$2*(8*6+8*2+6*2)\\\\=2*(48+16+12)\\=2*76\\=152$ ( $\text{cm}^2$ )

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2 years ago

X²+Y²=250 and XY=117<br> What are the values of X and Y?

yan [13]

Answer:

Step-by-step explanation:

$x^2+y^2=250\\\\x^2-2xy+y^2+2xy=250\\\\(x-y)^2=250-2xy\\\\(x-y)^2=250-2\cdot117\\\\ (x-y)^2=16\\\\x-y=4\qquad\qquad\vee\qquad \qquad x-y=-4\\\\x=4+y \qquad\qquad \vee\qquad\qquad x=-4+y\\\\(y+4)y=117\qquad\vee\qquad\quad (y-4)y=117\\\\y^2+4y-117=0\qquad\vee\qquad y^2-4y-117=0\\\\y=\dfrac{-4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\qquad\vee\qquad y=\dfrac{4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\\\\y=\dfrac{-4\pm\sqrt{16+468}}{2}\qquad\ \ \vee\qquad y=\dfrac{4\pm\sqrt{16+468}}{2}$

$y_1=\dfrac{-4-22}{2}\ ,\quad y_2=\dfrac{-4+22}{2}\ ,\quad y_3=\dfrac{4-22}{2}\ ,\quad y_4=\dfrac{4+22}{2}\\\\y_1=-13\ ,\qquad y_2=9\ ,\qquad\quad\qquad\ y_3=-9\ ,\qquad y_4=13\\\\x_{1,2}=4+y_{1,2}\qquad\qquad\qquad\qquad\qquad x_{3,4}=-4+y_{3,4}\\\\x_1=-9\ ,\qquad x_2=13\ ,\qquad\quad\qquad x_3=-13\ ,\qquad x_4=9$

6 0

4 years ago

A spinner is divided into eight equal-sized sections, numbered 1 to 8, inclusive. What is true about spinning the spinner one at

zvonat [6]

Answer with explanation:

The number of section into which Spinner is divided =Number from 1 to 8={1,2,3,4,5,6,7,8}

Now, Spinner is Spun once .

The Correct statement among the given statements are:

→1. S={1,2,3,4,5,6,7,8}

→2. If A is a subset of S, A could be {1,2,3} , because , {1,2,3}⊂{1,2,3,4,5,6,7,8}.

→5.If a subset A represents the complement of spinning an odd number, it’s sample space is A={2,,4,6,8}

A=odd number={1,3,5,7}

A's Complement=Even Number={2,4,6,8}

Why, not 3 and 4, because, The number of elements in the set is from 1 to 8, but the subset has three elements, 7,8, and 9.Element, 7 ∈ S, 8∈S, but , 9∈S.

Also, number less than 4, is equal to ={1,2,3}, but in the statement, it is given that,number less than 4, it’s sample space is A={1,2,3,4},which is Incorrect.

7 0

3 years ago

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