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3 years ago

12

an astronomer wishes to measure on a photograph the distance between the image of a certain star and three star other star image

s nearby, if 8 images are nearby how many choices of the three stars does he have?

1 answer:

Greeley [361]3 years ago

6 0

The astronomer has to choose 3 star images out of 8.This is a problem of combinations. This can be expressed as 8C3 i.e. combination of 8 objects taken 3 at a time.

$8C3= \frac{8!}{3!*5!}=56$

This means, if 8 images are nearby the astronomer will have 56 choices <span>of the three stars.</span>

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Use the following statements to find a compound statement for (p∨q)∧rp∨q)∧r p: 5<-3 q: All vertical angles are congruent. r:

guajiro [1.7K]

Given the compound statement <span>(p∨q)∧r
where: p: 5 < -3
q : All vertical angles are congruent.
r: 4x = 36, then x = 9.

Recall the in logic, '</span>∨' symbolises "or" while '∧' symbolises "and".

Therefore, the compound statement <span>(p∨q)∧r can be written as follows:
Either 5 < -3 or all vertical angles are congruent, and if 4x = 36, then x = 9.
</span>

4 0

3 years ago

What is x+x+x+x+x+x+x+x+x+x+b+b+b+b+3=

xz_007 [3.2K]

Answer:

10x+4b+3

Step-by-step explanation:

x+x+x+x+x+x+x+x+x+x+b+b+b+b+3

There are 10 x

4 b

and one number

10x+4b+3

8 0

2 years ago

Ben consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Ben's body decr

Airida [17]

Answer:

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

Step-by-step explanation:

After consuming the energy drink, the amount of caffeine in Ben's body decreases exponentially.

This means that the amount of caffeine after t hours is given by:

$A(t) = A(0)e^{-kt}$

In which A(0) is the initial amount and k is the decay rate, as a decimal.

The 10-hour decay factor for the number of mg of caffeine in Ben's body is 0.2722.

1 - 0.2722 = 0.7278, thus, $A(10) = 0.7278A(0)$ . We use this to find k.

$A(t) = A(0)e^{-kt}$

$0.7278A(0) = A(0)e^{-10k}$

$e^{-10k} = 0.7278$

$\ln{e^{-10k}} = \ln{0.7278}$

$-10k = \ln{0.7278}$

$k = -\frac{\ln{0.7278}}{10}$

$k = 0.03177289938
$

Then

$A(t) = A(0)e^{-0.03177289938t}$

What is the 5-hour growth/decay factor for the number of mg of caffeine in Ben's body?

We have to find find A(5), as a function of A(0). So

$A(5) = A(0)e^{-0.03177289938*5}$

$A(5) = 0.8531$

The decay factor is:

1 - 0.8531 = 0.1469

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

7 0

2 years ago

san4es73 [151]

Answer:

$9450

Step-by-step explanation:

We will use compound interest formula:

$F=P(1+r)^t$

Where

F is future amount [what we want to figure out]

P is present amount [9000]

r is rate of interest [since we want for 6 months, the annual interest divided by 2 is r. So r = 10/2 = 5% or 0.05]

t is the time [ the time period is for 6 months so t = 1 since we already converted the interest rate to 6 month chunk]

Putting in formula, we get:

$F=P(1+r)^t\\F=9000(1+0.05)^1\\F=9000(1.05)\\F=9450$

6 0

3 years ago

The quadratic formula gives which roots for the equation 2x^2 + x - 6 = 0?

Jet001 [13]

Answer:

The roots are $x = (\frac{3}{2}, -2)$ , which is given by option C.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

In this question, we have that:

$2x^2 + x - 6 = 0$

Which is a quadratic equation with $a = 2, b = 1, c = -6$ . So

$\Delta = 1^{2} - 4*2(-6) = 1 + 48 = 49$

$x_{1} = \frac{-1 + \sqrt{49}}{2*2} = \frac{6}{4} = \frac{3}{2}$

$x_{2} = \frac{-1 - \sqrt{49}}{2*2} = \frac{-8}{4} = -2$

So the roots are $x = (\frac{3}{2}, -2)$ , which is given by option C.

3 0

2 years ago