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2 years ago

14

What is the y-intercept of an equation

2 answers:

Alona [7]2 years ago

8 0

Answer:

The y interpit means the y-axis. you should've known this in 5th grade.

Step-by-step explanation:

krek1111 [17]2 years ago

4 0

Answer:

The point on a graph where a line crosses the Y-axis (The vertical line)

Step-by-step explanation:

You might be interested in

Find the perimeter of the polygon.

Liono4ka [1.6K]

Answer:

80

Step-by-step explanation:

The Two-Tangent Theorem states that if two tangent segments are drawn to one circle from the same external point, then they are congruent

13+13+9+9+12+12+6+6= 80

6 0

2 years ago

Commercial agents earn 5% of the cost of each product they sell. If an agent earns

kicyunya [14]

Based on the information the cost of the product is $40,000.

Given:

Percentage earn=5% or 0.05

Amount earned=$2,000

Let x represent the cost of the product

Hence:

Formulate

.05x = $2000

Divide both sides by .05

x=$2,000/.05

x=$40,000

Inconclusion the cost of the product is $40,000.

Learn more about cost here:brainly.com/question/824281

5 0

2 years ago

Read 2 more answers

What is the value of 5a when a=7 ?

Mariulka [41]

Answer:

35

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7 0

2 years ago

Read 2 more answers

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

Does the following represent the Transitive Property?

Sauron [17]

No it does not. If q=b and b=t then q=t is the correct answer.

7 0

3 years ago