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brilliants [131]
3 years ago
15

PLZ HELP WILL GIVE BRAINLIEST IF ANSWER CORRECTLY

Mathematics
1 answer:
lakkis [162]3 years ago
6 0

Answer:

3 = y

Step-by-step explanation:

y = 1/2 * 6x

y = 3x

-3y = x

-3y as a constant is (3y)

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Evaluate 2p-nm-4, if m=3, p=15, and<br> n=5.
BabaBlast [244]

Step-by-step explanation:

given m = 3, p = 15 and n = 5,

2p - nm - 4 \\  = 2(15) - (5)(3) - 4 \\  = 30 - 15 - 4 \\  = 15 - 4 \\  = 11

3 0
3 years ago
1 shirt and 3 sodas equal $30. How much does a shirt cost and how much does a soda cost?
gizmo_the_mogwai [7]

Answer:

I think 1 shirt costs $15 and each soda costs $5

Step-by-step explanation:

5*3=15

and 15+15=30

So the shirt has to be $15

6 0
4 years ago
Consider the quadratic function f(x) = x2 – 5x + 6.
Jobisdone [24]

Answer:

x^2 and -5x is a coefficients and 6 is a constant

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Draw the preimage of a triangle with coordinates T(2,1), U(0,-1) and V,(3,-3). Then use the following coordinates plane to compl
alekssr [168]
I don’t even know the answer sorry
3 0
3 years ago
Read 2 more answers
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
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