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tangare [24]
3 years ago
11

What is the place value for 7 in 0.673

Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
8 0
.07 or seven hundredths
lbvjy [14]3 years ago
3 0
The place value for 7 in 0.673 :

is 7 Hundredths

Hope this explains it.
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Solve the inequality 15t > 180
bagirrra123 [75]

Answer:

t>12

Step-by-step explanation:

t>180/15

then divide 180 by 15

and you gey t>12

4 0
3 years ago
Can anyone help me solve this
yawa3891 [41]

Answer:

a. $45.54; b. $51.42; c. See below

Step-by-step explanation:

a. 193 min

        Service charge = $40.00

1st 100 min @ 0.021 =      2.10

Next 93 min @0.037 = <u>    3.44 </u>

                        Total =  $45.54

b. 317 min

          Service charge = $40.00

  1st 100 min @ 0.021 =      2.10

2nd 100 min @ 0.037 =     3.70

        117 min @ 0.048 =   <u>   5.62</u>

                          Total =   $51.42

c. Piecewise function

The charge is  

  • $40.00 + 0.021t               if t ≤ 100
  • $42.10 + 0.037(t - 100)    if 100 < t ≤ 200
  • $45.80 + 0.048 (t - 200) if t > 200

which we can write like this:

f(t) =\begin{cases}40.00 + 0.021t & t \leq 100\\42.10 + 0.037(t-100) & 100 < t \leq200\\45.80 + 0.048(t-200) & t > 200\end{cases}

5 0
3 years ago
Netflix recommendations??​
faltersainse [42]

Answer:

Here wacth this chico bon bon very good show

Step-by-step explanation:

its good

4 0
3 years ago
Read 2 more answers
Describe the graph of the function.y = |x – 4| – 7
juin [17]
This is always ''interesting'' If you see an absolute value, you always need to deal with when it is zero:

(x-4)=0 ===> x=4,

so that now you have to plot 2 functions!

For x<= 4: what's inside the absolute value (x-4) is negative, right?, then let's make it +, by multiplying by -1:

|x-4| = -(x-4)=4-x
Then:

for x<=4, y = -x+4-7 = -x-3

for x=>4, (x-4) is positive, so no changes:

y= x-4-7 = x-11,

Now plot both lines. Pick up some x that are 4 or less, for y = -x-3, and some points that are 4 or greater, for y=x-11

In fact, only two points are necessary to draw a line, right? So if you want to go full speed, choose:

x=4 and x= 3 for y=-x-3

And just x=5 for y=x-11

The reason is that the absolute value is continuous, so x=4 works for both:

x=4===> y=-4-3 = -7

x==4 ====> y = 4-11=-7!

abs() usually have a cusp int he point where it is =0

Hope it helps, despite being this long!
3 0
3 years ago
Read 2 more answers
Consider the following function.
Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is f(x)=1+\frac{5}{x}-\frac{4}{x^2}

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, x=0 is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

f(x)=1+\frac{5}{x}-\frac{4}{x^2}

f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}

0=-\frac{5}{x^2}+\frac{8}{x^3}

0=-5x+8

5x=8

x=\frac{8}{5}

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}

f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3

Therefore, the function increases on the interval (0,\frac{8}{5}) and decreases on the interval (-\infty,0),(\frac{8}{5},\infty).

C) Since we determined that the slope is 0 when x=\frac{8}{5} from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}, meaning there's an extreme at the point (\frac{8}{5},\frac{41}{16}), but is it a maximum or minimum? To answer that, we will plug in x=\frac{8}{5} into the second derivative which is f''(x)=\frac{10}{x^3}-\frac{24}{x^4}. If f''(x)>0, then it's a minimum. If f''(x), then it's a maximum. If f''(x)=0, the test fails. So, f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}, which means (\frac{8}{5},\frac{41}{16}) is a local maximum.

D) Now set the second derivative equal to 0 and solve:

f''(x)=\frac{10}{x^3}-\frac{24}{x^4}

0=\frac{10}{x^3}-\frac{24}{x^4}

0=10x-24

-10x=-24

x=\frac{24}{10}

x=\frac{12}{5}

We then test where f''(x) is negative or positive by plugging in test values. I will use -1 and 3 to test this:

f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34, so the function is concave down on the interval (-\infty,0)\cup(0,\frac{12}{5})

f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0, so the function is concave up on the interval (\frac{12}{5},\infty)

The inflection point is where concavity changes, which can be determined by plugging in x=\frac{12}{5} into the original function, which would be f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}, or (\frac{12}{5},\frac{43}{18}).

E) See attached graph

5 0
3 years ago
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