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3 years ago

Rewrite as a logarithmic equation. e^(7)=y

Mathematics

1 answer:

jolli1 [7]3 years ago

5 0

E^7 = y --->(Add ln on both sides) lne^7 = ln y ---> (Bring down the power) (7)×lne = ln y ---> ( lne = 1 ) 7 = ln y

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I need help!!! Brainliest for anyone who answers correctly on both!!!

timama [110]

Answer:

The first answer: 390m The second answer: x = 7; m<1 = 17; m<2 = 73

Step-by-step explanation:

First problem:

It's asking for the distance he walks by walking around a rectangular box which is the perimeter.

Two of the sides will have length 115m and two will have length 80m. So this perimeter is 115 + 115 + 80 + 80 = 390m.

Second problem:

Complementary angles add to 90 degrees. So m<1 and m<2 added together will equal 90

-4x +45 + 7x + 24 = 90

3x + 69 = 90

3x = 21

x= 7.

Plug back x into both m<1 and m<2 to find the measure of the angles.

m<1 = -4(7) + 45 = 45-28 = 17

m<2 = 7(7) + 24 = 49 + 24 = 73

3 0

3 years ago

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→1 3x x − 1

Reika [66]

Answer:

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})$ , we are to evaluate it. To evaluate it, we will simply substitute x = 1 into the function since the variable x tends to 1.

$\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= (\dfrac{3(1)}{1-1} - \dfrac{3}{ln1})\\\\= \dfrac{3}{0} - \dfrac{3}{0}\\\\= \infty - \infty (ind)$

Since we got an indeterminate function, we will find the LCM of the function and solve again.

$= \lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= \lim_{x \to 1} \dfrac{3xlnx-3(x-1)}{(x-1)lnx}\\\\\\= \dfrac{3(1)ln(1)-3(1-1)}{(1-1)ln1}\\\\= \frac{3(0)-3(0)}{0(0)} \\\\= \frac{0}{0} (ind)$

Applying L'hospital rule;

$\frac{x}{y} = \lim_{x \to 1} \dfrac{d/dx(3xlnx-3(x-1))}{d/dx((x-1)lnx)}\\\\= \lim_{x \to 1} \dfrac{3x(\frac{1}{x})+ 3lnx-3)}{(x-1)\frac{1}{x} +lnx}\\\\= \lim_{x \to 1} \dfrac{3 + 3lnx-3}{(x-1)\frac{1}{x} +lnx}\\\\= \frac{3ln1}{(1-1)\frac{1}{1} +ln1}\\\\= \frac{0}{0} (ind)$

Applying L'hospital rule again;

$= \lim_{x \to 1} \dfrac{\frac{d}{dx} (3lnx)}{\frac{d}{dx} ((x-1)\frac{1}{x} +lnx)}\\\\= \lim_{x \to 1} \dfrac{\frac{3}{x} }{(x-1)\frac{-1}{x^2} + \frac{1}{x} +\frac{1}{x} }\\\\= \dfrac{\frac{3}{1} }{(1-1)\frac{-1}{1^2} + \frac{1}{1} +\frac{1}{1} }\\\\= \frac{3}{0(-1)+2}\\ \\= \frac{3}{2-0}\\ \\= 3/2$

<em>Hence the limit of the function is 3/2.</em>

7 0

3 years ago

Determine the value of x.

Readme [11.4K]

I think is (B) x= 2…

3 0

3 years ago

1+1 please answer correctly

Wewaii [24]

Answer:

2 is the answer

Step-by-step explanation:

easy thx for the pionts

5 0

3 years ago

Find Each Product Simplify<br> ( x ^ 2 + 9 ) ( x + 3 ) ( x - 3) HELP ME PLEASE!

Kisachek [45]

(x^2+9)(x+3)(x-3)=(x^2+9)(x^2-9)=x^4-81
Have a nice day!!! ^^

6 0

3 years ago