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3 years ago

Which is the first step in simplifying the expression?

Mathematics

2 answers:

Neko [114]3 years ago

6 0

Easy, multiply the expression by 2+divided by 2 over 3

In any other case, its D

Elenna [48]3 years ago

5 0

$\dfrac{3}{2 + \sqrt{2} }$

To rationalise the denominator, this is a standard step:
$= \dfrac{3}{2 + \sqrt{2} } \times \dfrac{2 - \sqrt{2}}{2 - \sqrt{2}}$

Answer: Option A

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Let me continue to complete the question, though it is not ask for

$= \dfrac{3(2 - \sqrt{2})}{(2 + \sqrt{2)}(2 - \sqrt{2}) }$

$= \dfrac{3(2 - \sqrt{2})}{4 - 2 }$

$= \dfrac{6 - 2\sqrt{2}}{2 }$

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What fraction of 16 is 12?

lisov135 [29]

Answer:

3/4

Step-by-step explanation:

7 0

3 years ago

Nate finishes his garden at 4:47 p. After pruning his roses for 18 minutes and watering his vegetables for 45 minutes. What time

3241004551 [841]

He finish his gardening at 5:50 p.m.

3 0

3 years ago

Solve for x:a(a²+b²)x²+b²x-a

m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

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

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}$

$\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}$

5 0

3 years ago

The sum of the areas of two circles is 8pie square meters. Find the length of a radius of each circle ifone of them is twice as

NeX [460]

Nice

area=pir^2
the areas are
pir^2+pi(2r)^2
they add to 8pi

8pi=pir^2+pi(2r)^2
undistribute pi
8pi=pi(r^2+(2r)^2)
divide both sides by pi
8=r^2+(2r)^2
expand
8=r^2+4r^2
add
8=5r^2
divide by 5
8/5=r^2
sqrt both sides
√(8/5)=r
double since it is double
2√(8/5) is radius of bigger aka (4√10)/5

radius=(4√10)/5

5 0

3 years ago

Mark all the statements that are true.

Artyom0805 [142]

Answer:

Option A

Option C

Step-by-step explanation:

A relationship is defined as a function if and only if each element of the "domain" set is assigned only one element of the "range" set. That is, there is only one output value y assigned to each input value x.

The relation x = 3 is not a function because there are infinite output values y, assigned to the same input element x.

(3, 2), (3, 5) (3, 9) (3,10000)

Then the option A is true

Option C is also true, the equation of the line shown is

$x = 3$

The rest of the options are false because x = 3 is not a function

3 0

3 years ago