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3 years ago

Evaluate the expression. If necessary, round to the nearest hundredth. lne^3x^2

2 answers:

7 0

<span>lne^3x^2

by the properties of logarithm

lna^x = xlna

so, the equation goes by

3x^2 lne; lne = 1

so the equation is 3x^2</span>

Naya [18.7K]3 years ago

5 0

For this case we have the following expression:
lne ^ 3x ^ 2
Rewriting we have:
ln (e ^ (3x ^ 2))
By properties of the natural and exponential logarithm function, we can
rewrite the expression as:
ln (e ^ (3x ^ 2)) = 3x ^ 2
Answer:
The equivalent expression is given by:
3x ^ 2

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Alex787 [66]

SOLUTION:

$Point \ (1,3)$

<h2>Explanation:</h2>

Let's use the following System of Linear Equations in Two Variables in order to solve this exercise:

$\begin{array}{c}(1)\\(2)\end{array}\left\{ \begin{array}{c}y=2x+1\\y=-4x+7\end{array}\right.$

So we can graph these lines using two points.

For eq (1):

$y=2x+1 \\ \\ If \ x=0 \ then \ y=2(0)+1 \therefore y=1 \\ \\ If \ y=0 \ then \ 0=2x+1 \therefore x=-\frac{1}{2} \\ \\ \\ Points: \\ \\ (0,1) \\ \\ (-\frac{1}{2},0)$

So graph a line that passes through these two points. This is the red line shown below.

For eq (2):

$y=-4x+7 \\ \\ If \ x=0 \ then \ y=-4(0)+7 \therefore y=7 \\ \\ If \ y=0 \ then \ 0=-4x+7 \therefore x=\frac{7}{4} \\ \\ \\ Points: \\ \\ (0,7) \\ \\ (\frac{7}{4},0)$

So graph a line that passes through these two points. This is the blue line shown below.

<em>The solution of this system of equation is the point of intersection, which is (1,3)</em>

<h2>Learn more:</h2>

Parallel lines: brainly.com/question/12169569

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3 years ago

4 cm<br>3 cm<br>x cm<br>area = 19 cm2

statuscvo [17]

Answer:

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3 years ago

Read 2 more answers

A rectangle is three centimeters.longer than it is wide. if its length were to be decreased by two centimeters, its area would d

Scorpion4ik [409]

Given/equations formed from text:
I: l=w+3
II: A=l*w
III: new area B=(l-2)*w
IV: B=A-30

substitute l in II with I to remove l:
II': A=l*w=
(w+3)*w=
w^2+3w

substitute l in III with I to remove l:
III': B=(l-2)*w=
(w+3-2)*w=
(w+1)*w=
w^2+w

substitute A and B from II' and III' into IV:
B=A-30
w^2+w=w^2+3w-30
30=2w
w=15

insert w=15 into I:
l=15+3=18
->
A) dimensions of full size rectangle are width 15 and length 18
dimensions of reduced size rectangle are width 15 and length 16

B) full size: 15*18=10*18+5*18=180+90=270cm^2
reduced size: 15*16=10*16+5*16=160+80=240cm^2

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4 years ago

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Step-by-step explanation:

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4 years ago

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Answer:

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Step-by-step explanation:

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3 years ago