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3 years ago

14

By how many degrees did the direction of the boat change when it made its first turn? Round to the nearest degree.

2 answers:

Vadim26 [7]3 years ago

6 0

The Boat Turned 39 Degrees

kkurt [141]3 years ago

6 0

Answer:

39 degrees is the correct answer

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Vlada [557]

We know that

When we have

$f(x)=ax^2 +bx+c$

Here ,

a is the leading coefficient

c is the constant term

so, we can compare it with

$f(x)=5x^2-7x+6$

so, we get

$a=5,c=6$

so,

leading coefficient is 5

constant term is 6......................Answer

4 0

3 years ago

craig went out to a restaurant for dinner. he tipped 16% of his bill, which brought the sum of the dinner bill plus the tip to $

olga nikolaevna [1]

The total cost was 116% of the original bill;
Original bill was;
$\frac{100}{116} *26.10$
= <span>$22.5</span>

5 0

3 years ago

B is 30 more than a. B is 6 times as many as a. What is the value of a and b ?

12345 [234]

Answer:

180 is the answer you multiple 6 times 30

4 0

3 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

An acute angle, θ, is in a right triangle such that sin of theta is equal to 3 over 8 period What is the value of cot θ?

Serggg [28]

We want to find the value of cot(θ) given that sin(θ) = 3/8 and θ is an angle in a right triangle, we will get:

cot(θ) = (√55)/3

So we know that θ is an acute angle in a right triangle, and we get:

sin(θ) = 3/8

Remember that:

sin(θ) = (opposite cathetus)/(hypotenuse)
hypotenuse = √( (opposite cathetus)^2 + (adjacent cathetus)^2)

Then we have:

opposite cathetus = 3

hypotenuse = 8 = √(3^2 + (adjacent cathetus)^2)

Now we can solve this for the adjacent cathetus, so we get:

adjacent cathetus = √(8^2 - 3^2) = √55

And we know that:

cot(θ) = (adjacent cathetus)/(opposite cathetus)

Then we get:

cot(θ) = (√55)/3

If you want to learn more, you can read:

brainly.com/question/15345177

8 0

3 years ago