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3 years ago

Simplify 1/2√40 A) √10 B) 2√10 C) 4√10

Mathematics

2 answers:

const2013 [10]3 years ago

7 0

$\frac{1}{2} \sqrt{ {2}^{2} } \times \sqrt{10}$
that is:
1/2 * 2 *√10

= √10 therefore Option A is correct

Len [333]3 years ago

3 0

It would br A)√10
If you want the work tell me.

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Find the length of 1 side of square if the area is 49 sq.cm method needed

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Answer:

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Step-by-step explanation:

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2 years ago

27x + 24y= 4.5 1.5x + y =0.225

Goryan [66]

Answer:

x = 0.1, y = 0.075

Step-by-step explanation:

Given the 2 equations

27x + 24y = 4.5 → (1)

1.5x + y = 0.225 → (2)

Multiplying (2) by - 24 and adding to (1) will eliminate the y- term

- 36x - 24y = - 5.4 → (3)

Add (1) and (3) term by term to eliminate y

- 9x = - 0.9 ( divide both sides by - 9 )

x = 0.1

Substitute x = 0.1 into either of the 2 equations and solve for y

Substituting into (2)

1.5(0.1) + y = 0.225

0.15 + y = 0.225 ( subtract 0.15 from both sides )

y = 0.075

3 0

3 years ago

Read 2 more answers

Luis and John deliver soft drinks to local convenience stores. Luis can complete the deliveries in 4 hours alone. John can compl

zaharov [31]

3 hours I think?
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8 0

3 years ago

Let u be an angle in the 4th quadrant with cos u=8/9 Then sin u=

lutik1710 [3]

To solve for the last side of the triangle, use the Pythagorean Theorem:
(8)^2 + x^2 = (9)^2
x = sqrt of 17
However, this is a NEGATIVE sqrt 17 because the terminal side is in quadrant 4, meaning that this side is under the X-axis and therefore negative.
Now that you know the side opposite of u in the triangle, do opposite/hypotenuse.
sin u = -(sqrt 17)/9

5 0

3 years ago

Solve the equation in the complex number system. x^2-12x+40=0

mylen [45]

solving the equation $x^2-12x+40=0$ we get $x=6+2i\,\,and\,\,x=6-2i$

Step-by-step explanation:

Solve the equation:

$x^2-12x+40=0$

We can solve using quadratic Formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b = -12, c = 40

Putting values:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(40)}}{2(1)}\\x=\frac{12\pm\sqrt{144-160}}{2}\\x=\frac{12\pm\sqrt{-16}}{2}\\We\,\,know\,\,\sqrt{-1}=i\\ x=\frac{12\pm4i}{2}\\ x=\frac{12+4i}{2}\,\,and\,\, x=\frac{12-4i}{2}\\ x=\frac{4(3+i)}{2}\,\,and\,\, x=\frac{4(3-i)}{2}\\x=2(3+i)\,\,and\,\,x=2(3-i)\\x=6+2i\,\,and\,\,x=6-2i$

So, solving the equation $x^2-12x+40=0$ we get $x=6+2i\,\,and\,\,x=6-2i$

Keywords: Solving quadratic equation

Learn more about Solving quadratic equation at:

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5 0

3 years ago