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3 years ago

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logan swam 750 meters in 1/3 hour mila swam 3/3 of logans distance in 1/4 hour what is logans average swimming speed? what is mi

las swimming speed? show work

1 answer:

user100 [1]3 years ago

3 0

Answer:

Logan's average swimming speed = 0.625 metres/second

Mia's average swimming speed0.833 metres/second

Step-by-step explanation:

Speed is obtained by dividing the distance covered by the time taken to cover such a distance.

<em>It will be necessary to make sure we are working in seconds for our time units.</em>

Logan's swimming time: 1/3 hours will be = 20 minutes = 1200 seconds

Mia's swimming time: 1/4 hours will be = 15 minutes = 900 seconds

Logan swam 750 metres. Logan's swimming speed will be

750metres / 1200 seconds = 0.625 metres/second

Mia's distance swam is 3/3 of Logan's. If we reduce the fraction 3/3 to its lowest term, we will have that it will be = 1/1.

1/1 X 750metres = 750 metres.

This means that Mia swam exactly the same distance as Logan.

Mia's swimming speed = 750 metres/ 900 seconds = 0.833 metres/second

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The general term for the sequence 2, 4, 8, 16, 32, . . . is

Nuetrik [128]

We are given

sequence is 2 , 4 , 8 ,16 , 32 , .....

Firstly , we will check whether it is geometric sequence

Checking geometric sequence:

We will find common ratio between successive terms

and then we check whether they are equal

r1=(second term)/(first term)

$r_1=\frac{4}{2}=2$

r2=(third term)/(second term)

$r_2=\frac{8}{4}=2$

r3=(fourth term)/(third term)

$r_3=\frac{16}{8}=2$

r4=(fifth term)/(fourth term)

$r_4=\frac{32}{16}=2$

we can see that all four ratios are same

$r_1=r_2=r_3=r_4=2$

so, this is geometric sequence

Calculation of general term:

We got

common ratio is

$r=2$

Let's assume

number of terms is n

first term is 2

$a_1=2$

now, we can use formula

$a_n=a_1 (r)^{n-1}$

we can plug values

and we get

$a_n=2(2)^{n-1}$

$a_n=2^1(2)^{n-1}$

$a_n=(2)^{n-1+1}$

$a_n=(2)^{n}$ ..................Answer

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3 years ago

Graph the line with slope – 2 passing through the point (-5,5).

34kurt

Answer:

y = -2x - 5

Step-by-step explanation:

y = -2x + b

- To find the y-intercept, plug the values of the variables.

5 = -2(-5) + b

5 = 10 + b

- Subtract 10 from both sides.

-5 = b

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2 years ago

A 5 pound bag of clay cost $2.65. At this rate how much would a 2 pound bag cost?

Nutka1998 [239]

2.65/5=.53(price per pound)
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3 years ago

9. A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

SSSSS [86.1K]

Answer:

Part 4) $r=84\ units$

Part 9) $sin(\theta)=-\frac{\sqrt{5}}{3}$

Part 10) $sin(\theta)=-\frac{9\sqrt{202}}{202}$

Step-by-step explanation:

Part 4) A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

we know that

The circumference of a circle subtends a central angle of 360 degrees

The circumference is equal to

$C=2\pi r$

using proportion

$\frac{2\pi r}{360^o}=\frac{56\pi}{120^o}$

simplify

$\frac{r}{180^o}=\frac{56}{120^o}$

solve for r

$r=\frac{56}{120^o}(180^o)$

$r=84\ units$

Part 9) Given cos(∅)=-2/3 and ∅ lies in Quadrant III. Find the exact value of sin(∅) in simplified form

Remember the trigonometric identity

$cos^2(\theta)+sin^2(\theta)=1$

we have

$cos(\theta)=-\frac{2}{3}$

substitute the given value

$(-\frac{2}{3})^2+sin^2(\theta)=1$

$\frac{4}{9}+sin^2(\theta)=1$

$sin^2(\theta)=1-\frac{4}{9}$

$sin^2(\theta)=\frac{5}{9}$

square root both sides

$sin(\theta)=\pm\frac{\sqrt{5}}{3}$

we know that

If ∅ lies in Quadrant III

then

The value of sin(∅) is negative

$sin(\theta)=-\frac{\sqrt{5}}{3}$

Part 10) The terminal side of ∅ passes through the point (11,-9). What is the exact value of sin(∅) in simplified form?

see the attached figure to better understand the problem

In the right triangle ABC of the figure

$sin(\theta)=\frac{BC}{AC}$

Find the length side AC applying the Pythagorean Theorem

$AC^2=AB^2+BC^2$

substitute the given values

$AC^2=11^2+9^2$

$AC^2=202$

$AC=\sqrt{202}\ units$

so

$sin(\theta)=\frac{9}{\sqrt{202}}$

simplify

$sin(\theta)=\frac{9\sqrt{202}}{202}$

Remember that

The point (11,-9) lies in Quadrant IV

then

The value of sin(∅) is negative

therefore

$sin(\theta)=-\frac{9\sqrt{202}}{202}$

5 0

3 years ago

Someone please help with 9th question !!!!

meriva

$LIMITS \: \: AND \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: DERIVATIVES$

Heya !

Check the attachment.
Hope it helps you :)

5 0

3 years ago