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3 years ago

Which logarithmic equation is equivalent to the exponential equation below? e^a=55

Mathematics

1 answer:

Volgvan3 years ago

7 0

Answer:

The answer is B.

e^a=55

Taking log on both sides,

lne^a = ln 55

a lne= ln55

a= ln55.

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Write the equation of a line that goes through point (0,1) and has a slope of 0

8090 [49]

Answer:

$1 = y$

Step-by-step explanation:

The y-intercept is [0, 1], so you draw a horizontal line through the line $y = 1.$ , which is considered a zero slope [rate of change].

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4 years ago

Peterkin park has a square fountain with a walkway aroind it. The fountain measures 12 feet on each side. The walkway is 3 1/2 f

Black_prince [1.1K]

Answer: 829.44 ft²

Step-by-step explanation:

Correction - The walkway is 31.2 feet wide.

To solve this, first find the area of the fountain:

Area of a square = Length * Width

= 12 * 12

= 144 ft²

The find the area of the walkway including the fountain:

= 31.2 * 31.2

= ‭973.44‬ ft²

To find the area of the walkway alone, subtract the area of the square from the area of the rectangle:

= 973.44 - 144

= 829.44 ft²

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3 years ago

Analysis of an accident scene indicates a car was traveling at a velocity of 69.5 mph (31.1 m/s) along the positive x-axis at th

STatiana [176]

We can find the acceleration via

${v_f}^2-{v_i}^2=2a\Delta x$

We have

$\left(5.20\dfrac{\rm m}{\rm s}\right)^2-\left(31.1\dfrac{\rm m}{\rm s}\right)^2=2a(115\,\mathrm m)$

$\implies\boxed{a=-4.09\dfrac{\rm m}{\mathrm s^2}}$

Then by definition of average acceleration,

$a_{\rm ave}=\dfrac{v_f-v_i}t$

so that

$-4.09\dfrac{\rm m}{\mathrm s^2}=\dfrac{5.20\frac{\rm m}{\rm s}-31.1\frac{\rm m}{\rm s}}t$

$\implies\boxed{t=6.33\,\mathrm s}$

We alternatively could have found the time without knowing the acceleration. Since acceleration is constant, the average velocity is

$v_{\rm ave}=\dfrac{x_f-x_i}t=\dfrac{v_f+v_i}2$

Then

$\dfrac{115\,\rm m}t=\dfrac{5.20\frac{\rm m}{\rm s}+31.1\frac{\rm m}{\rm s}}2$

$\implies\boxed{t=6.33\,\mathrm s}$

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3 years ago

Write the quadratic equation that has roots -1-rt2/3 and -1+rt2/3 if its coefficient with x^2 is equal to 1

weeeeeb [17]

The equation of the quadratic function is f(x) = x²+ 2/3x - 1/9

<h3>How to determine the quadratic equation?</h3>

From the question, the given parameters are:

Roots = (-1 - √2)/3 and (-1 + √2)/3

The quadratic equation is then calculated as

f(x) = The products of (x - roots)

Substitute the known values in the above equation

So, we have the following equation

$f(x) = (x - \frac{-1-\sqrt{2}}{3})(x - \frac{-1+\sqrt{2}}{3})$

This gives

$f(x) = (x + \frac{1+\sqrt{2}}{3})(x + \frac{1-\sqrt{2}}{3})$

Evaluate the products

$f(x) = (x^2 + \frac{1+\sqrt{2}}{3}x + \frac{1-\sqrt{2}}{3}x + (\frac{1-\sqrt{2}}{3})(\frac{1+\sqrt{2}}{3})$

Evaluate the like terms

$f(x) = x^2 + \frac{2}{3}x - \frac{1}{9}$

So, we have

f(x) = x²+ 2/3x - 1/9

Read more about quadratic equations at

brainly.com/question/1214333

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1 year ago

PLZ HELP Math show work 1. Graph the function: () = 3/2 − 4

MissTica

The answer is one in a half

7 0

3 years ago