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3 years ago

7

Is 10/3 an improper fraction

2 answers:

padilas [110]3 years ago

3 0

Yes, because 10 is greater than 3.

IceJOKER [234]3 years ago

3 0

Yes because the numerator is bigger than the denominator

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Simplify the expression 1,000,000 × 0.00062 in scientific notation and express your answer in scientific notation.

Morgarella [4.7K]

The answer is 620. you can use a calculator for these types of questions

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3 years ago

What is 4/3 divided by 5/3

Deffense [45]

4/5 because if you divide it would be 12/15 so I simplified

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4 years ago

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A vector has initial point at (3,-5) and terminal point at (-2,3). what are the characteristics of this vector?

Aleksandr [31]

Answer:

D

Step-by-step explanation:

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3 years ago

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The triangles are similar, solve for x.<br> N+3<br> 8

Nonamiya [84]

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5

Step-by-step explanation:

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3 years ago

Square root of the quantity x minus 2 end quantity plus 8 equals x

Butoxors [25]

ANSWER

x=11

EXPLANATION

We want to solve the equation:

$\sqrt{x - 2} + 8 = x$

Add -8 to both sides:

$\sqrt{x - 2} = x - 8$

Square both sides of the equation to get:

$(\sqrt{x - 2} )^{2} =( x - 8)^{2}$

$x - 2 = {x}^{2} - 16x + 64$

Rewrite in the form ax² +bx+c=0

${x}^{2} - 16x - x + 64 + 2 = 0$

${x}^{2} - 17x + 66 = 0$

Factor to get,

$(x - 11)(x - 6) = 0$

This implies that,

$x = 11 \: or \: x = 6$

Checking for extraneous solutions.

When x=6, the original equation becomes,

$\sqrt{6- 2} + 8 = 6$

$\sqrt{4 } + 8 = 6$

2+8= 6 is not true.

x=6 is an extraneous solution.

When x=11, the original equation becomes,

$\sqrt{11- 2} + 8 = 11$

$\sqrt{9} + 8 = 11$

$3 + 8 = 11$

This is true, hence x=11 is the only solution.

5 0

3 years ago