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2 years ago

Derek was 4 years 2 months old when he joined school. Today he is 11 years 1 month old. For how long has he been in school?

Mathematics

1 answer:

lora16 [44]2 years ago

6 0

Answer:

Step-by-step explanation:

first of all,we should do this:

1 year is 12 month

amd 4 year is 48 month(12×4)

then we must add 2 months (wich is given) to 48

He was 50 months old when he started school

and we should find how old is he now

11 ×12=132months

132+1=133months

133-50=83 ( he has been 83 months in school)

let's turn it to years 6 years 11 months

( for finding years you should divide 83 to 12)

hope it helps

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Evaluate: 6-(2/3)^2 A. 17/3 B. 52/3 C. 49/9 D. 50/9

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The correct answer is: "Option [D]".

Step-by-step explanation:

Hi student, let me help you out!

....................................................................................................................................

Let's use the acronym PEMDAS. With the help of this little acronym, we will not make mistakes in the Order of Operations! :)

$\dag\textsf{Acronym \: PEMDAS}$

P=Parentheses,

E=Exponents,

M=Multiplication,

D=Division,

A=Addition,

S=Subtraction.

Now let's start evaluating our expression, which is $\mathsf{6-(\cfrac{2}{3})^2}$

According to PEMDAS, the operation that we should perform is "E-Exponents".

Notice that we have a fraction raised to a power. When this happens, we raise both the numerator (2 in this case) and the denominator (3 in this case) to that power, which is 2. After this we obtain $\mathsf{6-\cfrac{4}{9}}$ .

See, we raised both the numerator and the denominator to the power of 2.

Now what we should do is subtract fractions.

Note that 6 and -4/9 have unlike denominators. First, let's write 6 as a fraction: $\mathrm{\cfrac{6}{1}-\cfrac{4}{9}}$ . Now let's multiply the denominator and the numerator of the first fraction times 9: $\mathrm{\cfrac{54}{9}-\cfrac{4}{9}}$ .

See, now the fractions have the same denominator. All we should do now is subtract the numerators: $\mathrm{\cfrac{50}{9}}$ .