Please answer quickly

A scientist wants to find

maria [59]

Your answer is 8 your answer

3 0

2 years ago

Read 2 more answers

Which number is an irrational number

allochka39001 [22]

Hi there! The answer is C.

An irrational number is a number that cannot be written as a fraction.

A, B and D can be written as a fraction.
A. 3.5 = 3 1 / 2
B. square root of 36 = 6 = 6 / 1
D. 37 / 100 is already a fraction.

Hence, the only number that cannot be written as a fraction is the square root of 48.

$\sqrt{48} = 6.928...$

The answer is C: the square root of 48 is an irrational number.

7 0

4 years ago

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This question has me confused so help please!!

atroni [7]

Answer:

Δ MNO ≅ Δ RST

Step-by-step explanation:

In naming the congruent triangles follow the same order, that is

M is the vertex with 79° and R is the vertex with 79°

MN = 21 and RS = 21

These are the first 2 letter names of the triangles, that is MN and RS

O and T are the third, thus

Δ MNO ≅ Δ RST

8 0

3 years ago

Please help!!! i really need some help on this :( I'll make you Brainliest!!!

shusha [124]

78+92=170
170/2=85

his current average is 85% so id go with C knowing that if his next quiz he got at least a 70, his average would be 80%

5 0

3 years ago

Hi, so I know the answer to this problem (now that I got it wrong) but I'm not quite sure why I was wrong, help?

IgorC [24]

Answer:

$-2x^2-x$ . Your answer was wrong because "x-squared" terms didn't cancel.

Step-by-step explanation:

To solve this problem, set up an equation. We know something is supposed to be added to the expression $2x^2+2x$ , and the result should be x. So:

$2x^2+2x+(\text{ ? })=x$

We want to solve for the question mark... the unknown thing that we're adding to the original expression, in order to get x.

It is uncommon to put question marks in equations to represent quantities. Usually we use a letter. Since x is already being used in the equation, we should pick something else ... we could use "y".

$2x^2+2x+(\text{ } y \text{ })=x$

...or just...

$2x^2+2x+y=x$

Algebra allows us to solve the equation and find out what "y" is equivalent to.

To solve, we want to get the "y" by itself. To do so, we try to eliminate the other "terms" from the left side of the equation.

Understanding "terms" & "like terms"

Terms

"Terms" in an equation are either a number multiplied to other things, or just a single number that isn't multiplied to anything else.

For example, the various terms in our equation above are

$2x^2$ , $2x$ , $y$ , $x$

You might ask why the last things, which don't have a number, are considered terms.

Remember that multiplying by 1 doesn't change anything, so we could imagine each of the last two terms as being 1 times the letter.

So, we can rewrite our equation:

$2x^2+2x+1y=1x$

Like terms

"Like terms" are terms where the "other stuff the numbers are multiplied to" is the same, so for instance, the $2x$ and the $1x$ are like terms. They are like terms because, the "other stuff" that the numbers are multiplied to are "x" for both terms. Note that $2x$ and $2x^2$ are not "like terms" because the "stuff" is different:

$x$ is different than $x^2$

"Like terms" are important because only like terms can be "combined" into a single simplified term.

Solving equations

To solve an equation, we isolate what we're solving for, y, by disconnecting the other terms from it, and simplify.

Starting with subtracting 2x from both sides of the equation:

$2x^2+2x+1y=1x\\(2x^2+2x+1y)-2x=(1x)-2x$

Subtraction is the same as "adding a negative":

$2x^2+2x+1y+(-2x)=1x+(-2x)$

Since all terms are now connected by addition, we can add in any order we want (because of the Commutative Property of Addition), and we can combine like terms.

Thinking just about the number parts, since $1+(-2)=-1$ , then $1x+(-2x)=-1x$ .

Returning to our main equation, the right side simplifies:

$2x^2+2x+1y+(-2x)=1x+(-2x)\\2x^2+2x+1y+(-2x)=-1x$

On the left side: $2x$ and $-2x$ are like terms.

Fact: $2+(-2)=0$

So, $2x+(-2x)=0x$

Since anything times zero is just zero, $0x=0$ . Furthermore, adding zero to anything doesn't change it. So when the $2x$ and $-2x$ terms on the left side of our main equation are combined, they "disappear" (While we talked through are a lot of rules/steps to justify why that works, it is common to omit those justifications, and to just combine those like terms and make them disappear.)

So, $2x^2+2x+1y+(-2x)=-1x$ simplifies to:

$2x^2+1y=-1x$

Similarly for the $2x^2$ term, we subtract from both sides:

$2x^2+1y=-1x\\(2x^2+1y)-2x^2=(-1x)-2x^2\\2x^2+1y+(-2x^2)=-1x+(-2x^2)$

Combining like terms on the left, they disappear.

$1y=-1x+(-2x^2)$

There are no like terms on the right.

Since the two terms on the right are added together, we can use the commutative property of addition to rearrange:

$1y=-2x^2+(-1x)$

Addition of a negative can turn back into subtraction, and simplify multiplication by 1.

$y=-2x^2-x$

Remembering we chose "y" as the unknown thing we wanted to know, that's why the "correct answer" is what it is.

Verifying an answer

Verifying can double check an answer, and helps explain why the answer you chose doesn't work.

To verify an answer, the original statement said add something to the expression and get a result of "x". So, let's see if the "correct answer" does:

$2x^2+2x+(\text{ } ? \text{ })\\2x^2+2x+(-2x^2-x)\\2x^2+2x+(-2x^2-1x)\\2x^2+2x+(-2x^2)+(-1x)$

Combining the "x-squared" terms, completely cancels...

$2x+(-1x)$

Combining the "x" terms, and simplifying...

$1x\\x$

So it works.

Why isn't the answer what you chose:

$2x^2+2x+(\text{ } ? \text{ })\\2x^2+2x+(-x^2-x)\\2x^2+2x+(-1x^2-1x)\\2x^2+2x+(-1x^2)+(-1x)$

Combining the x-squared terms, things don't completely cancel...

$1x^2+2x+(-1x)$

Combining the x terms...

$1x^2+1x\\x^2+x$

So adding the answer that you chose to the expression would not give a result of "x", which is why it is "wrong"

5 0

2 years ago