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3 years ago

Someone tell me where everyone is going right please !!

Mathematics

2 answers:

ivolga24 [154]3 years ago

8 0

Answer:

Step-by-step explanation:

( 2 , 3 ) , ( 6 , - 5) , ( - 1 , 3 ) - is a function

( 1 , 9 ) , ( -3 , - 2 ) , ( 1 , - 4 ) - is a function

( 7 , - 4 ) , ( 0 , 9 ) , ( 2 , - 2 ) - is a function

( 0 , 3 ) , ( 0 , 7 ) , ( 4 , 0 ) - is a function

( -6 , 5 ) , ( -5 , 6 ) , ( 8 , 2 ) - is a function

marysya [2.9K]3 years ago

5 0

1,3and 5 are functions

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Really need help with math question.

JulsSmile [24]

The opposite angles of a cyclic quadrilateral add up to 180 degrees , so

2x + 3 + 4x + 3 = 180
6x = 180 - 6

6x = 174
x = 29

so m<C = 2(29) + 1 = 59 degrees

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3 years ago

70 is 30% of what number?

ExtremeBDS [4]

Answer: 233.33.

Step-by-step explanation: Consider marking this answer as brainliest if it helped you out.

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3 years ago

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An isosceles triangle with a 35 angle and an 110 angle

aalyn [17]

The remaining angle is 35 degrees. This makes the triangle an isosclese triangle as well.

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3 years ago

A rectangle has dimensions 3x-1 and x+6. Write an expression for the area of the rectangle as a polynomial in standard form

WITCHER [35]

Answer:

the area product is simply (3x+5).(x+7 which can multiply if desired to obtain highlight3x/2+ 26x + 35.

You can use Completion of the Square on the trinomial product to put this trinomial into standard form. You would want this form to be like (x-h)2 +k.

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3 years ago

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