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3 years ago

Need help please, thanks!

1 answer:

4 0

Given:
1 inch : 18 kilometers

Let us use the ratio and proportion method:

a:b = c:d where ad = bc

0.8 inches is 14.4 kilometers

1 : 18 = 0.8 : x
x = 0.8 * 18
x = 14.4 km

1.4 inches is 25.2 kilometers

1: 18 = 1.4 : x
x = 18 * 1.4
x = 25.2 km

2.1 inches is 37.8 kilometers

1 : 18 = 2.1 : x
x = 18 * 2.1
x = 37.8 km

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According to a recent study, the carapace length for adult males of a certain species of tarantula are normally distributed with

seraphim [82]

Idk tbh sorry bye ok hbye

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

3 years ago

suppose x and y vary inversely and y=10 when x=4. write a function that models the variation then find y when x =90

ankoles [38]

X = k/y
4 = k/10
k = 4 x 10 = 40
Thus, x = 40 / y

90 = 40 / y
y = 40/90 = 4/9

4 0

3 years ago

Need help with this math

Oksi-84 [34.3K]

Answer:
B. 6 + √26
Step-by-step explanation:
1. Make a right triangle using the line that goes from office to supermarket.
2. Use pythagorean thereom to find the distance of the longest side.
3. FInd the distance of the line segment going from supermarket to Marias House.
4. Add them together to get 6 + √26.

6 0

3 years ago

Read 2 more answers

dino has 14 coins and 2 one dollar bills which describes the relationships between the coins and the dollar bills

kkurt [141]

Answer:

they are both money

Step-by-step explanation:

that's why they relate, they're both currency

6 0

3 years ago

Read 2 more answers