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3 years ago

If 2 sq rt 2x = 2 sq rt 3, what is the value of x?

Mathematics

1 answer:

Arturiano [62]3 years ago

5 0

Answer:

<u>The correct answer is x = 3/2</u>

Step-by-step explanation:

2 √2x = 2√3, what is the value of x?

Let's solve for x, this way:

2 √2x = 2√3

√2x = √3 (Dividing by 2 at both sides)

(√2x) ² = (√3) ² (Raising both sides at the 2nd power)

2x = 3

x = 3/2 (Dividing by 2 at both sides)

<u>The correct answer is x = 3/2</u>

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15+3x=45 solve the equation

inna [77]

Answer:

x= 10

Step-by-step explanation:

45-15=30

30/3=10

3 0

4 years ago

I need help on 25. I know the answer is 2a but I need to see the work. Please help I’ve tried so many possibilities.

valkas [14]

a(2a-3(1-a))+5(a-a^2)

distribute the inner most parenthesis

a( 2a -3+3a))+5(a-a^2)

combine like terms

a(5a-3) +5(a-a^2)

distribute both sets of parenthesis

5a^2 -3a +5a-5a^2

combine like terms

5a^2-5a^2 -3a+5a

7 0

3 years ago

Read 2 more answers

Suppose A and B are two matrices with the same dimensions. Explain how to find A + B and B - B.

SCORPION-xisa [38]

A+B: the matrix with same dimension a A and B, the element in row i and column j is: Aij+Bij

Same for A-B, but with Aij-Bij

6 0

3 years ago

Nicole used 3/8 of her ribbon to wrap a present if she used 6 feet of ribbon for the present how much ribbon did Nicole have at

Bas_tet [7]

Nicole used only a fraction of her ribbon to wrap the present

the length of ribbon she used is 6 feeet

so lets name the length of the ribbon she had originally as X

the fraction of X she used was - 3/8 of X

so we can write the following equation

$\frac{3}{8} of X = 6$

when of function is used we multiply

$\frac{3}{8}*X = 6$

3X = 6*8

3X = 48

X = 16

therefore length of the original ribbon piece Nicole had was 16 feet

7 0

3 years ago

Read 2 more answers

If x = 4 tan(θ), find sec(θ) in terms of x

prisoha [69]

Answer:

$\sec \theta = \pm \frac{ \sqrt{ {x}^{2} + 16} }{4}$

Step-by-step explanation:

$\because \: x = 4 \tan \theta \\ \therefore\frac{x}{4} = \tan \theta....(1) \\ \\ \because \: { \sec}^{2} \theta = 1 + { \tan}^{2} \theta \\ \therefore \: \sec \theta = \pm\sqrt{1 + { \tan}^{2} \theta } \\ \therefore \: \sec \theta = \pm\sqrt{1 + { \bigg( \frac{x}{4} \bigg)}^{2} } \\ \therefore \: \sec \theta = \pm\sqrt{1 + { \frac{x^{2}}{16} } } \\ \therefore \: \sec \theta = \pm\sqrt{{ \frac{16 + x^{2}}{16} } } \\ \therefore \: \sec \theta = \pm \frac{ \sqrt{ {x}^{2} + 16} }{4}$

3 0

3 years ago