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4 years ago

Miguel buys textbook with a listed price of $81. Miguel uses

Mathematics

2 answers:

77julia77 [94]4 years ago

8 0

Answer:

Step-by-step explanation: First, multiply the $81 by the 10% (81 x .10 = 8.1). Next, use the answer from step one and subtract it from the $81 (81 - 8.1 = 72.90) . $72.90 will be your final answer.

zimovet [89]4 years ago

6 0

Answer: $72.90

Step-by-step explanation:

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The Answer is D. 1/3

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3 years ago

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Mila [183]

(-3) + (-5) + 9

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2 years ago

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3 years ago

A ball is thrown into the air with a velocity of 34 ft/s. It's height in feet after t seconds is given by y=34t-26(t)^2. Find th

kkurt [141]

Answer:

The average velocity of the ball at the given time interval is -122.3 ft/s

Step-by-step explanation:

Given;

velocity of the ball, v = 34 ft/s

height of the ball, y = 34t - 26t²

initial time, t₀ = 3 seconds

final time, t = 3 + 0.01 = 3.01 seconds

At t = 3 s

y(3) = 34(3) - 26(3)² = -132

The average velocity of the ball in ft/s is given as;

$V_{avg} = \frac{y(3.01)-y(3)}{3.01 -3}\\\\V_{avg} = \frac{34(3.01)-26(3.01)^2-y(3)}{3.01 -3}\\\\V_{avg} = \frac{-133.223-y(3)}{0.01}\\\\V_{avg} = \frac{-133.223-(-132)}{0.01}\\\\V_{avg} =\frac{-1.223}{0.01}\\\\V_{avg} = -122.3 \ ft/s$

Therefore, the average velocity of the ball at the given time interval is -122.3 ft/s

8 0

3 years ago

For what values of x is log base 0.8 (x+4)>log base 0.4 (x+4)<br>Thank you!

Radda [10]

We are seeking the solution of the inequality:

$\displaystyle{ \log_{0.8}(x+4)\ \textgreater \ \log_{0.4}(x+4)$ .

We recall that a log function $f(x)=\log_b(x)$ is either increasing or decreasing:

i) it is increasing if b>1,

ii) it is decreasing if 0<b<1.

Consider the functions $\displaystyle{ \log_{0.8}(x)$ and $\displaystyle{ \log_{0.4}(x)$ .

The graphs of these functions both meet at x=1 (clearly), and after 1 they are both negative. So from 0 to 1 one of them is larger for all x, and from 1 to infinity the other is larger. (Being strictly decreasing, their graphs can only intersect once.)

We can check for a certain convenient point, for example x=0.8:

$\displaystyle{ \log_{0.8}(0.8)=1$ and

$\displaystyle{ \log_{0.4}(0.8)=\log_{0.4}(0.4\cdot 2)=\log_{0.4}(0.4)+\log_{0.4}(\cdot 2)=1+\log_{0.4}(2)$ .

Now, $\displaystyle{ \log_{0.4}(2)$ is negative since we already explained that for x>1 both functions were negative. This means that

$\displaystyle{ \log_{0.8}(0.8)\ \textgreater \ \log_{0.4}(0.8)$ , and since $0.8\in (0, 1)$ , then this is the interval where $\displaystyle{ \log_{0.8}(x)\ \textgreater \ \log_{0.4}(x)$ .

So, now considering the functions $\displaystyle{ \log_{0.8}(x+4)$ and $\displaystyle{ \log_{0.4}(x+4)$ , we see that

x+4 must be in the interval (0,1), so we solve:

0<x+4<1, which yields -4<x<-3 after we subtract by 4.

Answer: (-4, -3). Attached is the graph generated using Desmos.

7 0

3 years ago