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DochEvi [55]
3 years ago
8

Simplify. Use the ^ to show an exponent and the / for the fraction bar.

Mathematics
1 answer:
patriot [66]3 years ago
4 0
X exponent 3 / 2y

hope this helps!
You might be interested in
Find the expression x = -7 x2 -1 / x+4
ELEN [110]
Please confirm the question is x = (-7x^2 - 1) / (x + 4)

if that is the case, then

x * (x + 4) = -7x^2 - 1

x^2 + 4x = -7x^2 -1

8x^2 + 4x + 1 = 0

which does not have any real solution


Edit - much easier question to evaluate the expression (x^2 -1) / (x + 4) when x=-7

(-7^2 - 1) / (-7 + 4)

= (49 -1 ) / -3

= 48 / -3

= -16


7 0
3 years ago
Hey can you guys help me please help
yawa3891 [41]
The triangle on the left area is 18 because the area of a triangle is 1/2 b times h so 12 times 3 times 1/2 equals 18.you then cut a triangle from the quadrilateral to make into a triangle.The triangle is the same height has other triangle so its area is 18.Since you took three for base of left and right triangle you subtract 6 from 1 to make it your base of 8 and your height is 12 and 12 tikmes  8 is 96 so 18+18+96=132 so our answer is 132.
4 0
3 years ago
Is f(x)=5x and g(x)=x/5 inverse of each other?
Nana76 [90]

Answer:

Yes it is!

Step-by-step explanation:

3 0
3 years ago
There are 6 gray keys for every 11 black keys. Also, the number of black keys is 2 less than twice the number of gray keys. Use
emmainna [20.7K]

\text{ Answer: There are }12\text{ gray keys.}\\\\\text{ there are }22\text{ black keys.}

Explanation:

Since we have given that

There are 6 gray keys for every 11 black keys .

So, the ratio will be 6:11

Let the number of gray keys be 6x

Let the number of black keys be 11x

Now, according to question,

11x=2\times 2x-2\\\\11x=12x-2\\\\11x-12x=-2\\\\-x=-2\\\\x=2

so,

\text{ there are }6x=6\times 2=12\text{ gray keys.}\\\\\text{ there are }11x=11\times 2=22\text{ black keys.}

8 0
3 years ago
Prove: Every point of S=(0, 1) is an interior point of S.
myrzilka [38]

Answer:

Proved

Step-by-step explanation:

To prove that every point in the open interval (0,1) is an interior point of S

This we can prove by contradiction method.

Let, if possible c be a point in the interval which is not an interior point.

Then c has a neighbourhood which contains atleast one point not in (0,1)

Let d be the point which is in neighbourhood of c but not in S(0,1)

Then the points between c and d would be either in (0,1) or not in (0,1)

If out of all points say d1,d2..... we find that dn is a point which is in (0,1) and dn+1 is not in (0,1) however large n is.

Then we find that dn is a boundary point of S

But since S is an open interval there is no boundary point hence we get a contradiction. Our assumption was wrong.

Every point of S=(0, 1) is an interior point of S.

3 0
3 years ago
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