MATHEMATICS HIGH SCHOOL

An object is traveling around a circle with a radius of 10 cm. If in 20 seconds a central angle of 1/3 radian is swept out, what is the linear speed of the object?

Answers

Answer 1

Answer:

The linear speed of the object is $(1)/(6)$

The Radius, r = 10 cm
The Fraction of Radius swept out = $(1)/(3)$
The distance = $10 / (1)/(3)$
The distance covered = $(10)/(3)$

The linear speed :

Radius swept out ÷ time taken
$(10)/(3) * (1)/(20) = (1)/(6)$

The linear speed of the object is $(1)/(6)$

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Answer 2

Answer: Firstly, we will find the distance covered by the object in the 20 seconds:

$\text{angle}(in~radians)=\frac{\text{arch}}{\text{radius}}\Longrightarrow (1)/(3)=(d)/(10~cm)\iff d=(10)/(3)~cm$

We must know that $v=(d)/(\Delta t)$ . So:

$v=(d)/(\Delta t)\Longrightarrow v=((10)/(3)~cm)/(20~s)\iff \boxed{v=(1)/(6)~cm/s}$

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Find the domain and range of the INVERSE of the given function.
Square root x-3

Answers

$f(x)=√(x-3)\n y=√(x-3)\n y^2=x-3\n x=y^2+3\n f^(-1)(x)=x^2+3\n D:x\in\matxbb{R}\n R:y\in\langle3,\infty)$

A bag contains 10 red marbles, 7 blue marbles, and 3 green marbles. You randomly select a marble from the bag, record the color, and then put that marble back in the bag. If you were to do this 200 times, how many times would you expect to pull out a blue marble?

Answers

Answer:

Step-by-step explanation:

We have 7 blue marbles, and 20 total marbles. As a result, if we randomly chose one marble from the bag, the probability would be (number of favorable outcomes)/(number of total outcomes) = 7/20.

This means that for each 20 trials, we would expect to pull out a blue marble 7 times, as the probability does not change over time (because we put the marbles back, there are the same amount of each marble). Over 200 trials, we can simply make it out of 200 instead of 20. We can thus multiply 7/20 by 10/10 (10/10 is equal to 1, keeping the proportions equal) to get 70/200, so we can expect 70 blue marbles out of 200.

Sam took his family to the zoo. An adult’s ticket is two timesthe cost of a child’s ticket. The total cost for two adults’ tickets and
three children’s tickets was $28. How much do the tickets cost?

Answers

cost of child ticket = x
cost of adult ticket = 2x
2(2x) + 3(x) = 28
4x + 3x = 28
7x = 28
x = 28/7
x = 4
Child ticket cost 4
Adult ticket cost 8

I am confused on numbers 25 and 29, the instructions are at the top.

Answers

#25 is fairly simple. Plug in -4 and 3 into the equation, and the extraneous root will be the one that does not work.
$√(12-(-4)) = √(16) = \frac{+}{}4$
Extraneous root in this case is positive four since +4≠-4
$√(12-3) = √(9) = \frac{+}{}3$
In this case it's negative 3, since -3≠3

#29 can be turned into a quadratic equation.
$x= √(2x+3)$
Square both sides to get
$x^(2)=2x+3$
Then bring the 2x+3 to the other side, setting the quadratic equal to zero.
$x^(2)-2x-3=0$
Factor to find that it's equivalent to
(x-3)(x+1)=0
Therefore x is equal to positive 3 and negative 1. Plug both back into the original equation. Whichever does not work is the extraneous root, and the answer is the one that does.
$x= √(2x+3)$
$3= √(2(3)+3)$
$3= √(9)$
Extraneous root would be negative 3.

$-1= √(2(-1)+3)$
$-1= √(1)$
Extraneous root would be positive 1.

Your answers are positive 3 and negative 1.
Extraneous roots are negative 3 and positive 1.

7. Numericals(a) A box has length 50 cm, breadth 20 cm and height 10 cm, then find
(i) its surface area of base;
(ii) its volume
(1000 cm², 100000 cm3)

Answers

Answer:

i = 1000cm2

ii = 100000cm3

Step-by-step explanation:

for q i

surface area of base = l* b

= 50* 20

1000cm^2

for q 2

volume l*b*h

= 50*20*10

= 100000cm^3

hope it will be helpfull

The hands of a clock form the same angle at various times of the day. For example, the angle formed at 2:00 is congruent to the angle formed at 10:00. If a clock has a diameter of 1 ft, what is the distance along the edge of the clock from the minute hand to the hour hand at 2:00? Give your answer in exacts.

Answers

it is 2 inches long this is because of the length of the hands .

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