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3 years ago

Jerry was 6 years old when his mother was 30. Now she is twice his age. How old is Jerry

Mathematics

1 answer:

Viktor [21]3 years ago

4 0

X=Jerry's age
y=his mother's age

y-x=24
y=2x

therefore x=24 and y=48

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There are 48 DVDs of comedies and dramas on the shelf. If 25% of the DVDs are comedies, how many are dramas?

Mazyrski [523]

There are 36 drama DVD's

<em><u>Solution:</u></em>

Given that there are 48 DVD's of comedies and dramas on shelf

25 % of DVD's are comedies

To find: number of drama DVD's

Total number of dvd = 48

Number of comedy dvd = 25 % of 48

Let us find 25 % of 48

$\rightarrow 25 \% \text{ of } 48\\\\\rightarrow 25 \% \times 48\\\\\rightarrow \frac{25}{100} \times 48\\\\\rightarrow 0.25 \times 48 = 12$

Thus, number of comedy dvd = 12

Thus we get,

number of drama DVD's = Total number of dvd - number of comedy dvd

number of drama dvd = 48 - 12 = 36

Thus there are 36 drama dvd's

4 0

3 years ago

I'm having trouble with this, it's geometry. please help and break it down if you can

Travka [436]

Check the picture below.

5 0

3 years ago

The National Aeronautics and Space Administration (NASA) purchased a new solar-powered battery guaranteed to have a failure rate

Reptile [31]

Answer:

The probability that all three batteries would fail is given as:

= 0.000125

Step-by-step explanation:

Given that the probability of the new battery failing = 1/20 or 0.05,

Therefore, the probability that all three batteries would fail is given as:

= Probability of battery A failing * Probability of battery B failing * Probability of battery C failing

= P(A)*P(B)*P(C)

= 0.05 * 0.05 * 0.05

= 0.000125

5 0

3 years ago

How much is one can of peaches if the price for three is 2.99?

BabaBlast [244]

If the price for three of them is $2.99, then you divide that by three to find the price for one can. This would equal .9966666667, which could probably be rounded to 99 cents or just a dollar per can. Hope that helps.

7 0

3 years ago

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

3 0

3 years ago