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3 years ago

6

The table shows the results of a survey of seventh-grade students in the lunch-line

1 answer:

Anni [7]3 years ago

5 0

Answer:

D

Look at the picture. :)

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What is (-10,8) in point- slope form?

nata0808 [166]

There can't be a point-slope form until you have TWO points.
A single point doesn't have any other form.
There are an infinite number of lines that all go through a single point.
Every one of them has a different slope, and a different point-slope form.

7 0

3 years ago

Which expression is equivalent to 6(4(6x-4y)-5) ?

iren2701 [21]

Answer:

$144x - 96y - 30$

Step-by-step explanation:

<u>Step 1: Distribute </u>

$6(4(6x - 4y) - 5)$

$6((4 * 6x) + (4 * -4y) - 5)$

$6(24x - 16y - 5)$

<u>Step 2: Distribute again </u>

<u /> $(6 * 24x) + (6 * -16y) + (6 * -5)$

$144x - 96y - 30$

Answer: $144x - 96y - 30$

8 0

3 years ago

It takes Emmanuel 20 minutes to walk to school. It takes Liah 80% as long to walk to school. How long does it take Liaqh to walk

77julia77 [94]

Answer:

16 minutes

Step-by-step explanation:

80% of 20 is

80/100 * 20 = 16

6 0

3 years ago

Explain how to find the axis of symmetry for each function<br> h(x)<br> The graph goes under h(x)

AleksAgata [21]

Find the minimum point on the parabola, and take the x component of the coordinate. Then, put that number into this equation: x = (whatever the number was) so in this case the axis (or line) of symmetry is x = 1

3 0

3 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

4 years ago