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masya89 [10]
3 years ago
13

What is 10 times 10000000

Mathematics
2 answers:
tatyana61 [14]3 years ago
7 0
100000000 is the answer
Aneli [31]3 years ago
5 0
100000000

10 times 10 million equals 100,000,000
think about how 10 times ten equals 100.



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Write an equation for the parabola that satisfies
givi [52]

Answer:

Step-by-step explanation:

y = -3(x-3)^2 + 4

4 0
3 years ago
Which number line represents the solution set for the inequality 2x – 6 ≥ 6(x – 2) + 8?
vovangra [49]

Answer: x≤ -0.5

Step-by-step explanation:  Graph the two lines or solve for x:

2x – 6 ≥ 6(x – 2) + 8

2x – 6 ≥ 6x – 12 + 8

2x – 6 ≥ 6x – 4

-4x  ≥  2

  x  ≤  -1/2  [Inequality is reversed when multiplying or dividing by a negative number]

The number line is anything less than or equal to - 1/2

Or one can plot each side of the inequality and note the interception point (x = -(1/2)).

4 0
3 years ago
Evaluate the expression when y=-4.<br> y²-7y+2
harkovskaia [24]

Answer:

Here is your answer

Hope it helps

6 0
3 years ago
Consider functions of the form f(x)=a^x for various values of a. In particular, choose a sequence of values of a that converges
sleet_krkn [62]

Answer:

A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.

Step-by-step explanation:

A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):

f(x)=a^{x}\\ f'(x)=a^xln(a)

The process will help us to understand what is happening, at first we rewrite the function:

f(x)=a^x\\ f(x)=e^{ln(a^x)}\\ f(x)=e^{xln(a)}\\

And then, we use the chain rule to differentiate:

f'(x)=e^{xln(a)}ln(a)\\ f'(x)=a^xln(a)

Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know  that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).

In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):

if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b)  (and below we explain why this happen)

Considering that ln(a) is a growing function and ln(e)=1, we have:

if a<e<b, then ln(a)< 1 <ln(b)

if a<e, then aˣln(a)<aˣ

if e<b, then bˣ<bˣln(b)

And because eˣ is defined to be the same as its derivative, the cases above results in the following

if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)

if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )

if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )

but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)

3 0
3 years ago
A square + b square =
dmitriy555 [2]

c square


its making me write 20 more characters

ignore this

the answer is c squared

                                         

8 0
3 years ago
Read 2 more answers
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