Math 10 Provincial Exam?

Have any of you guys heard of these, well I have one of these, only students of BC, have to complete these, If any of you are smart in math, could you help me with the following questions # 24, 25, 60. For some reason I cant copy and past the question. So heres the link to the exam

http://www.bced.gov.bc.ca/exams/search/grade10/english/sampl e/exam/0801ma_sample_exam.pdf

the answers for the questions are

#60)c

#24)c

#25)d

please help, i will really apreciate it! 😊

ok the questions are


# 60)The point (2,-1) is on the circumference of a circle with centre (-4,3). Determine the area of the circle.

Answer= 163.4


#24) If f(x)=-2x+3 and g(x)=1x/2-8, which statement is true?

a)f(-2)-g(-2)=-2
b)f(-2)-g(-2)=2
c)f(-2)+g(-2)=-2
d)f(-2)+g(-2)=2

Answer=C

Originally posted by: Bolly*_Crazed

ok the questions are


# 60)The point (2,-1) is on the circumference of a circle with centre (-4,3). Determine the area of the circle.

Answer= 163.4

i don't understand it

because i think the radius is 6 and if you put it in the formula.. the answer comes out wrong

do you know how to figure out the radius? cuz i even drew it =/

Lol sorry dear, i suck in maths 😕

Mathematics has always been my favourite subject.
24. If f(x)= –2x + 3 and g(x)= x/2 - 8 , which statement is true?
A. f(-2) - g(-2) = -2
B. f(-2) - g(-2) = 2
C. f(-2) + g(-2) = -2
D. f(-2)+ g(-2) = 2

Solution: -
f(x)= –2x + 3
Therefore, f(-2) = -2 * (-2) + 3 = 4 + 3 = 7

g(x) = x/2 - 8
Therefore, g(-2) = (-2)/2 - 8 = -1 - 8 = -9

f(-2) - g(-2) = 7 - (-9) = 7+ 9 = 16
Both A and B are wrong.

f(-2) + g(-2) = 7 + (-9) = 7 - 9 = -2
So, C is correct.

Ans: C

25. If g(x)= 2x + 2 and f(x)= 4x - 1, evaluate g(-3/sqrt(2)) - f (-1/sqrt(2))

A. -5 * sqrt(2)+ 1
B. -5 * sqrt(2)+ 3
C. -sqrt(2) + 1
D. -sqrt(2) + 3

Solution: -
g(x) = 2x + 2
Substituting x = -3/sqrt(2), we get
g(-3/sqrt(2)) = 2 * (-3/sqrt(2)) + 2
= -3 * (2/sqrt(2) + 2
= -3 * sqrt(2) + 2 ----------------------------------(1)

f(x)= 4x - 1
Substituting x = -1/sqrt(2), we get
f (-1/sqrt(2)) = 4 * (-1/sqrt(2)) - 1
= -2 * (2/sqrt(2)) - 1
= -2 * sqrt(2) - 1 -----------------------------(2)

From equations (1) and (2),

g(-3/sqrt(2)) - f (-1/sqrt(2)) = -3 * sqrt(2) + 2 - (-2 * sqrt(2) - 1)

= -3 * sqrt(2) + 2 + 2 * sqrt(2) + 1
= -3 * sqrt(2) + 2 * sqrt(2) + 2 + 1
= -sqrt(2) + 3

Ans: D

60. The point (2, -1) is on the circumference of a circle with centre (-4, 3). Determine the
area of the circle.
A. 25.1
B. 43.3
C. 163.4
D. 182.2

Solution: -
Let r be the radius of the circle.
Then r = distance between centre and a point on the circumference
Or, r = distance between (-4, 3) and (2, -1)
Or, r^2 = square of the distance between (-4, 3) and (2, -1)
(Note: - here ^2 stands for square.)
Or, r^2 = (-4-2)^2 + (3-(-1)))^2
= (-6)^2 + 4^2
= 36 + 16
= 52
Area of the circle = PI * r^2
= (22/7) * 52
= 163.4

Ans: C

thanks hun, i wouldve really apreciated your help, but my exam was on June 23 😊

Sorry, I saw this thread very late. In future, please let me know whenever you have any problem with Maths or Physics. Post the questions in any thread. If I do not reply, this means I did not see that question. In that case, just PM me.

Originally posted by: akhl

Sorry, I saw this thread very late. In future, please let me know whenever you have any problem with Maths or Physics. Post the questions in any thread. If I do not reply, this means I did not see that question. In that case, just PM me.

Oh wow! thanks hun 😛