Answer:
she had $60 before she went for shopping
Step-by-step explanation:
PLZ MARK BRAINLIEST
Let x represent the amount of money that Victoria had before she went for shopping.
Victoria spent one-fourth or her birthday money on clothes. It means that the amount she spent on shopping is 1/4 × x = x/4. Amount that she was having left would be x - x/4 = 3x/4
She received another 25$ a week later. The amount that she is having at this point will be 3x/4 + 25
If she has a total of 70$ now, it means that
3x/4 + 25 = 70
Multiplying through by 4
3x + 100 = 280
3x ,= 280 - 100 = 180
x = 180/3 = 60
Answer:
D) y=-4x+2
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(-2-6)/(1-(-1))
m=-8/(1+1)
m=-8/2
m=-4
y-y1=m(x-x1)
y-6=-4(x-(-1))
y-6=-4(x+1)
y=-4x-4+6
y=-4x+2
Answer:
57.5 mi
Step-by-step explanation:
1 in = 5 mi
11.5 × 5 = 57.5 mi
18. If f(x)=[xsin πx] {where [x] denotes greatest integer function}, then f(x) is:
since x denotes the greatest integers which could the negative or the positive values, also x has a domain of all real numbers, and has no discontinuous point, then x is continuous in (-1,0).
Answer: B]
20. Given that g(x)=1/(x^2+x-1) and f(x)=1/(x-3), then to evaluate the discontinuous point in g(f(x)) we consider the denominator of g(x) and f(x). g(x) has no discontinuous point while f(x) is continuous at all points but x=3. Hence we shall say that g(f(x)) will also be discontinuous at x=3. Hence the answer is:
C] 3
21. Given that f(x)=[tan² x] where [.] is greatest integer function, from this we can see that tan x is continuous at all points apart from the point 180x+90, where x=0,1,2,3....
This implies that since some points are not continuous, then the limit does not exist.
Answer is:
A]
Answer:
The length of mid-segment 
units
Step-by-step explanation:
Given:
Length of 
units
is the mid-segment which means the line joining the midpoints of two sides of triangle.
By mid-segment theorem the mid-segment of a triangle is parallel to the third side and half of the length of third side.
So, we can write:
∴ units
∴The length of mid-segment units
