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3 years ago

In a class of 30 pupils, there are 18 girls. what percentage of the pupils are boys

Mathematics

1 answer:

vovikov84 [41]3 years ago

6 0

Answer:

40%

Step-by-step explanation:

Express the boys as a fraction of the whole class then multiply by 100%

number of boys in class = 30 - 18 = 12

percentage of boys = $\frac{12}{30}$ × 100% = 40%

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mel-nik [20]

Answer

Step-by-step explanation:

You need to use cosine because the side length needed is adjacent and you have the hypotenuse.Set up a proportion of "cos23 = x/13" and solve by cross multiplying. Then just round.

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3 years ago

The diameter of a circle is 4cm. Find it area to nearewt tenth.

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Step-by-step explanation:

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2 years ago

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aliina [53]

The answer is Answer:

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3 years ago

7 x-3(4 x-8)<6 x+12-9 x

kari74 [83]

Answer:

x > 6

Step-by-step explanation:

1.) Distribute - 3 through the parenthesis

7x-3(4x-8)<6x+12-9x

7x-12x+24<6x+12-9x

2.) Then collect like terms

7x-12x+24<6x+12-9x

7x-12x+24<-3x+12

7x-12x+24<-3x+12

-5x+24<-3x+12

3.) Move the <u>variable</u> to the left and change the sign

-5x+24<-3x+12

-5x+3x+24<12

4.) Move the <u>constant</u> to the right and change it's sign

-5x+3x+24<12

-5x+3x<12-24

5.) Collect like terms

-5x+3x<12-24

-2x<12-24

6.) Calculate the difference

-2x<-12

7.) Divide both sides by -2 and flip the inequality sign

-2x<-12

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2 years ago

Use the properties of limits to help decide whether the limit exists. If the limit exists, find its value. ModifyingBelow lim W

Sergeu [11.5K]

Answer:

The value of given limit problem is 0.

Step-by-step explanation:

The given limit problem is

$lim_{x\rightarrow \infty}\dfrac{6x^3+5x-7}{6x^4-4x^3-9}$

We need to find the value of given limit problem.

Divide the numerator and denominator by the leading term of the denominator, i.e., $x^4$

$lim_{x\rightarrow \infty}\dfrac{\frac{6x^3+5x-7}{x^4}}{\frac{6x^4-4x^3-9}{x^4}}$

$lim_{x\rightarrow \infty}\dfrac{\frac{6}{x}+\frac{5}{x^3}-\frac{7}{x^4}}{6-\frac{4}{x}-\frac{9}{x^4}}$

Apply limit.

$\dfrac{\frac{6}{ \infty}+\frac{5}{ \infty}-\frac{7}{ \infty}}{6-\frac{4}{ \infty}-\frac{9}{ \infty}}$

We know that $\frac{1}{\infty}=0$ .

$\dfrac{0+0-0}{6-0-0}$

$\dfrac{0}{6}$

$0$

Hence, the value of given limit is 0.

8 0

3 years ago