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Veronika [31]
3 years ago
8

What is the simplest form of (–4v + 7)(3v – 5)?

Mathematics
1 answer:
Gennadij [26K]3 years ago
8 0
D.

(-4v*3v) + (-4v*-5) + (3v*7) + (7*-5)
-12v^2 + 20v + 21v -35
-12v^2 + 41v -35
You might be interested in
What are the possible solutions to the equation: x-12 = sqrt x+44
Ainat [17]

possible solutions for equation  x-12 = \sqrt{x+44} is  x=5 , x=20

<u>Step-by-step explanation:</u>

Here we have to find  the possible solutions to the equation: x-12 = sqrt x+44 .Let's find out:

⇒ x-12 = \sqrt{x+44}

⇒ (x-12)^2 = (\sqrt{x+44})^2

⇒ (x^2+144-24x) =x+44

⇒ (x^2+144-24x) -(x+44) =0

⇒ (x^2+100-25x) =0

⇒ x^2-25x+100 =0

⇒ x^2-20x-5x+100 =0

⇒ x(x-20-5(x-20) =0

⇒ (x-5)(x-20)=0

⇒ x-5=0 , x-20=0\\

⇒ x=5 , x=20

Therefore , possible solutions for equation  x-12 = \sqrt{x+44} is  x=5 , x=20

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
4 years ago
Can anyone please solve this and tell me if it's correct please
NikAS [45]

Answer:

Q1. 15
Q2. You're correct
Q3. 3.60
Q4. 2  Correct

Step-by-step explanation:

3 0
2 years ago
Circle each number below which is greater than its square.
Elenna [48]

Answer:

2/3 and -3 are correct hope it helps

4 0
3 years ago
Please add an explanation too ​
allsm [11]

Answer:

21

Step-by-step explanation:

21.2132034

6 0
3 years ago
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