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4 years ago

6

If f(x)=square root x-3, which inequality can be used to find the domain of f(x)? A square root x-3 =>0 B x-3=>0 C square

root x-3<=0 D x-3<=0

2 answers:

dangina [55]4 years ago

6 0

The answer is x-3=>0 i just took the quiz

goldenfox [79]4 years ago

5 0

we have

$f(x)=\sqrt{x-3}$

we know that

The term in the root square can not be a negative number

so

$(x-3)\geq0\\x\geq3$

the domain is the interval----------> [3,∞)

therefore

<u>the answer is the option </u>

B (x-3)=>0

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Simplify the following

den301095 [7]

Answer:

1) 11 $\sqrt{3}$

2) 2 $\sqrt{2}$

3) $20\sqrt{3} + 15\sqrt{2}$

4) $53 + 12\sqrt{10}$

5) -2

6) $7\sqrt{2} - 5\sqrt{3}$

Step-by-step explanation:

1) 2 $\sqrt{12}$ + 3 $\sqrt{48}$ - $\sqrt{75}$

=(2 × 2 $\sqrt{3}$ )+ (3 × 4 $\sqrt{3}$ ) - 5 $\sqrt{3}$

= 4 $\sqrt{3}$ + 12 $\sqrt{3}$ - 5 $\sqrt{3}$

= 11 $\sqrt{3}$

2) 4 $\sqrt{8}$ -2 $\sqrt{98}$ + $\sqrt{128}$

= (4 × 2 $\sqrt{2}$ ) - (2 × 7 $\sqrt{2}$ ) + 8 $\sqrt{2}$

= 8 $\sqrt{2}$ - 14 $\sqrt{2}$ +8 $\sqrt{2}$

= 2 $\sqrt{2}$

3) 5 $\sqrt{12\\}$ - 3 $\sqrt{18} + 4 \sqrt{72} +2\sqrt{75}$

= 5× $2\sqrt{3}$ - 3× $3\sqrt{2}$ + 4× $6\sqrt{2}$ + 2× $5\sqrt{3}$

= $10\sqrt{3} - 9\sqrt{2} +24\sqrt{2} +10\sqrt{3}$

= $20\sqrt{3} + 15\sqrt{2}$

4) $(2\sqrt{2} + 3\sqrt{5} )^{2}$

= $8 + 12\sqrt{10} + 45$

= $53 + 12\sqrt{10}$

5) $(1+\sqrt{3} ) (1-\sqrt{3} )$

= $1 - 3$

= -2

6) $(2\sqrt{6} -1) (\sqrt{3} -\sqrt{2} )$

= $2\sqrt{18}-2\sqrt{12} -\sqrt{3} +\sqrt{2}$

= 2× $3\sqrt{2}$ - 2× $2\sqrt{3}$ - $\sqrt{3} + \sqrt{2}$

= $6\sqrt{2} - 4\sqrt{3} -\sqrt{3} +\sqrt{2}$

= $7\sqrt{2} - 5\sqrt{3}$

Hope the working out is clear and will help you. :)

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3 years ago

Read 2 more answers

Solve the inequality 3x - 5 < 4.

Pachacha [2.7K]

Answer:

x < 3

Step-by-step explanation:

Since 3x - 5 < 4 then we can transpose the inequality to make x the subject of the equation:

If 3x - 5 < 4

⇒ 3x < 4 + 5

⇒ x < 9 ÷ 3

⇒ x < 3

8 0

3 years ago

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Answer:

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4 years ago

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Gnoma [55]

Answer:

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Step-by-step explanation:

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3 years ago

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Over the last three evenings, Donna received a total of 129 phone calls at the call center. The third evening, she received 3 ti

lakkis [162]

Answer:

Number of calls received in 1st evening = 24

Number of calls received in 2nd evening = 33

Number of calls received in 3rd evening = 72

Step-by-step explanation:

Let

number of calls received in first evening be "x"

number of calls received in 2nd evening be "y"

number of calls received in 3rd evening be "z"

Total, in 3 evenings, we have 129 phone calls, so we can write:

x + y + z = 129

3rd evening, 3 times as first evening. We can write:

z = 3x

2nd evening, 9 MORE THAN FIRST. So we can write:

y = 9 + x

Now we replace 2nd and 3rd equation in 1st equation to get:

x + y + z = 129

x + 9 + x + 3x = 129

Now, first, we solve for x:

$x + 9 + x + 3x = 129\\5x+9=129\\5x=120\\x=24$

Solving for y:

y = 9 + x

y = 9 + 24

y = 33

Solving for z:

z = 3x

z = 3(24)

z = 72

Thus,

Number of calls received in 1st evening = 24

Number of calls received in 2nd evening = 33

Number of calls received in 3rd evening = 72

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