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Papessa [141]
2 years ago
6

Find the area of 12m and 8m​

Mathematics
1 answer:
SCORPION-xisa [38]2 years ago
4 0

Answer:

Hiya!!

12m * 8m = 96m

Step-by-step explanation:

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Determinati numerele de trei cifre care,suprimand cifra zecilor, obtinem un numar de 13 ori mai mic.
lilavasa [31]

Answer:

English please

Step-by-step explanation:

6 0
2 years ago
15 pts. Prove that the function f from R to (0, oo) is bijective if - f(x)=x2 if r- Hint: each piece of the function helps to "c
solniwko [45]

Answer with explanation:

Given the function f from R  to (0,\infty)

f: R\rightarrow(0,\infty)

-f(x)=x^2

To prove that  the function is objective from R to  (0,\infty)

Proof:

f:(0,\infty )\rightarrow(0,\infty)

When we prove the function is bijective then we proves that function is one-one and onto.

First we prove that function is one-one

Let f(x_1)=f(x_2)

(x_1)^2=(x_2)^2

Cancel power on both side then we get

x_1=x_2

Hence, the function is one-one on domain [tex[(0,\infty)[/tex].

Now , we prove that function is onto function.

Let - f(x)=y

Then we get y=x^2

x=\sqrt y

The value of y is taken from (0,\infty)

Therefore, we can find pre image  for every value of y.

Hence, the function is onto function on domain (0,\infty)

Therefore, the given f:R\rightarrow(0.\infty) is bijective function on (0,\infty) not on  whole domain  R .

Hence, proved.

3 0
3 years ago
Find the equivalent expressions for (x + 7)(x + 1).
Natali5045456 [20]

Answer:

The answers should be

x^2 + 8x + 7

x(x + 8) +7

Step-by-step explanation:

(x + 7)(x + 1)

x(x) + x(1) + 7(x) + 7(1)

x^2 + x + 7x + 7

x^2 + 8x + 7

5 0
3 years ago
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Answer:

Call of duty. warzone=free

Step-by-step explanation:

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2 years ago
How to evaluate the limit
anzhelika [568]
\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}

Both the numerator and denominator are continuous at x=2, which means the quotient rule for limits applies:

\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2

Perhaps you meant to write that x\to-2 instead? In that case, you would have

\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5
4 0
3 years ago
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