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3 years ago

What is the slope of the line on the graph? Enter your answer in the box.

Mathematics

1 answer:

marishachu [46]3 years ago

5 0

Answer:

-2/6

Step-by-step explanation:

rise/run

how much you go up/how much you go to the side

the line is going to the left making it negative

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A large open meadow near your city is shrinking in size because developers have started building new homes there.

valentinak56 [21]

Answer:

$t=-10\log 0.32$

Explanation:

<u>Equation:</u>

$A=1,562.5\times 10^{(-0.1t)}$

This is a decaying exponential function, because the exponent is negative.

Then, the larger the value of t (number of months), the smaller A (the number of hectares).

The factor 1,562.5 is the initial number of hecares, when t = 0.

You must find the value of t when A equals 500, thus solve the equation:

$500=1,562.5\times 10^{(-0.1t)$

Divide both sides by 1,562.5

$\dfrac{500}{1,562.5}=10^{(-0.1t)}$

$0.32=10^{(-0.1t)}$

Take base-10 logarithms on both sides:

$\log 0.32=-0.1t$

$t=-\dfrac{\log 0.32}{0.1}\\\\\\t=-10\log 0.32$

Note that the logarithm of a number decimal number is negative number, thus t will be positive. When you compute that, it is about 4.9 months.

8 0

3 years ago

the waiter places a bowl of soup in front of lacy in a counterclockwise direction she passes the soup 90° the person receiving t

dybincka [34]

haifia is the answer did it on test

5 0

3 years ago

Which two shapes can be classified as parallelograms?

Deffense [45]

Actually there are three shapes that can be classified as a parallelogram.

Square, Rectangle, and Rhombus

3 0

3 years ago

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HELP ME PLEASE IS IMPORTANT AND ASAP!

zysi [14]

Answer:

maximun (0,5)

Step-by-step explanation:

6 0

3 years ago

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Use the algebraic procedure explained in section 8.9 in your book to find the derivative of f(x)=1/x. Use h for the small number

Triss [41]

Answer:

By definition, the derivative of f(x) is

$lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Let's use the definition for $f(x)=\frac{1}{x}$

$lim_{h\rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{x-(x+h)}{x(x+h)}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{(-1)h}{x^2+xh}}{h}=\\lim_{h\rightarrow 0} \frac{(-1)h}{h(x^2+xh)}=\\lim_{h\rightarrow 0} \frac{-1}{x^2+xh)}=\frac{-1}{x^2+x*0}=\frac{-1}{x^2}$

Then, $f'(x)=\frac{-1}{x^2}$

7 0

4 years ago