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2 years ago

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

Mathematics

2 answers:

Komok [63]2 years ago

5 0

Answer:

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: $\mathrm{\cfrac{50}{9}}$ .

∴, the answer is Option D.

Hope this helped you out, ask in comments if any queries arise.

Best Wishes!

$\star\bigstar\underline{\overline{\overline{\underline{\textsf{Reach \: far. Aim \: high. Dream \: big.}}}}}\bigstar\star$

$\underline{\rule{300}{5}}$

Arada [10]2 years ago

5 0

Calculate 2/3, to the power of 3, therefore

$\bf{\left(\dfrac{2}{3}\right)^{2}=\dfrac{2\times2}{3\times3}=\dfrac{4}{9} }$

$\bf{6-\dfrac{4}{9} }$

Convert 6 to the fraction 54/9.

$\bf{\dfrac{54}{9}-\dfrac{4}{9} }$

Since 54/9 and 4/9 have the same denominator, join their numerators to subtract them.

$\bf{\dfrac{54-4}{9} \ \ \to \ \ \ Subtract }$

Subtract 4 from 54 to get 50.

$\bf{\dfrac{50}{9} \ \ \to \ \ \ Answer }$

Therefore, the correct alternative is "D".

$\red{\boxed{\green{\boxed{\boldsymbol{\purple{Pisces04}}}}}}$

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yKpoI14uk [10]

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3 years ago

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If 30g of sugar is needed to make 4 cakes how much grams of sugar is needed to make 7 cakes

arsen [322]

There's two ways to do this.

The first way:
Work out the amount of sugar needed to make one cake, and then multiply that by 7.

30 grams ÷ 4 cakes = 7.5 grams for 1 cake.
7.5 grams × 7 cakes = 52.5 grams for 7 cakes.

The second way:
Work out the ratio of sugar:cake by doing 7 ÷ 4, and then multiplying that value by 30.

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3 years ago

Gin Rummy and are holding 10 cards in your hand from a standarddeck of 52 cards. Find the probability t

zepelin [54]

Here is the correct question

Gin Rummy and are holding 10 cards in your hand from a standard deck of 52 cards.

a) How many ways can you select three kings and four 6's from the deck, the remaining cards not kings or 6's

b) Find the probability that three kings and four 6's are being held by you from the deck.

Answer:

a) 52976

b) $\mathbf{3.34866744 \times 10^{-6}}$

Step-by-step explanation:

So from the information above,

no of ways three kings can be selected and four 6's from the deck without the remaining cards being kings or 6's can be expressed as:

no of ways to select 3 kings from 4 kings is $(^4_3)$

no of ways to select 4 6's from (4 6's) is $(^4_4)$

no of ways to select the remaining 3 cards from (52 - 4 - 4) = 44 cards is $(^{44}_3)$

Mathematically, multiplying the three together we have:

$= (^4_3) (^4_4) (^{44}_3)$

$= \dfrac{4!}{3!(4-3)!} \times \dfrac{4!}{4!(4-4)!} \times \dfrac{44!}{3!(44-3)!}$

$= 4 \times 1 \times 13244$

= 52976

To find the probability that you are holding three kings and four 6's from the deck, the remaining cards not kings or 6's can be calculated as follows;

To do this, we need to first know the no of ways we can select 10 cards out of the pack of 52 cards

i.e

$(^{52}_{10})$

= $\dfrac{52!}{10!(52-10)!}$

= $\dfrac{52!}{10!(42)!}$

= 1.58200242 × 10¹⁰

Now, to find the probability that you are holding three kings and four 6's from the deck, the remaining cards not kings or 6's :

we will need to divide the number of ways we can select three kings and four 6's from the deck by the no of ways we can select 10 cards out of the pack of 52 cards.

Mathematically; we have,

$=\dfrac{52976}{1.58200242 \times 10^{10}}$