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3 years ago

8x7+4x2-11-7x4 can you explain please

1 answer:

8 0

Use PEMDAS. Parenthesis. Exponents. Multiplication. Division. Addition. Subtraction. there are no parenthesis so that doesn't matter. there are no exponents. but there is multiplication. 8×7 = 56. so now the equation is 56+4×2-11-7×4. you still have multiplication so now we do 4×2 which is 8. the equation is now 56+8-11-7×4. you still have multiplication so now do 7×4 which is 28. the equation is now 56+8-11-28. there is no multiplication. or division. next is addition. 56+8 which is 64. now the equation is 64-11-28. next you do subtraction. 64-11 which is 53. now the equation is 53-28 which is 25. 25 is the answer. I hope this helped ☺️

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How many time 2/3 can go into 16<br><br> Explanation pls!!

tensa zangetsu [6.8K]

Answer:

24 times

Step-by-step explanation:

16 divided by 2/3 is 24

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2 years ago

What two numbers multiply to get 1440 but add to get 78

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The answer is 68, Which should be correct

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3 years ago

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What is the probability that 2 randomly selected months have 31 days?

aniked [119]

Answer:

31.8%

Step-by-step explanation:There are 7 months that have 31 days so,

The odds the first selected month has 31 days is 7/12

The next month there are only 6 months left with 31 days of only 11 months to choose from

so the odds for the second are 6/11

To calculate the odds they BOTH have 31 months you multiply the two odds:

(7/12) x (6/11) = 42/132 = .318 so 31.8%

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3 years ago

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How is the graph of y=3(x+1)^2 related to its parent function y=x^2

creativ13 [48]

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y = 3(x + 1)^2

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3 years ago

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I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu

ra1l [238]

The imaginary unit $i$ belongs to the set of complex numbers, denoted by $\mathbb C$ . These numbers take the form $a+bi$ , where $a,b$ are any real numbers.

The set of real numbers, $\mathbb R$ , is a subset of $\mathbb C$ , where each number in $\mathbb R$ can be obtained by taking $b=0$ and letting $a$ be any real number.

But any number in $\mathbb C$ with non-zero imaginary part is not a real number. This includes $i$ .

"is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because $2=2+0i$ .

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising $i$ to the $i$ -th power. Since $i=e^{i\pi/2}$ , we have

$i^i=\left(e^{i\pi/2}\right)^i=e^{i^2\pi/2}=e^{-\pi/2}\approx0.2079$

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3 years ago