Answer:
NPOM ≅ VUTS
OPNM ≅ TUVS
Step-by-step explanation: Given: two quadrilaterals having verticals P, N, O,M and S,T,V,U are congruent, where, OM is congruent or equal to TS and in quadrilaterals NPOM and VUTS-
since, the condition and, side UV=side OM follow for the above quadrilateral. (According to the figure)
then we can say according to the property of quadrilateral, their corresponding sides must be congruent. so they are congruent.
Answer:
The correct answer for this question is this one:
If the x position of the vertex for f(x) is k then :
1. f(k) => 4x+3=k, x = (k-3)/4 = (k/4)-(3/4)
2. x = k+3
3. x = (k+1)/2 = (k/2)-(1/2)
4. x = 2(k+2) = 2k+4
5. x = k+7
6. x = k-3
7. x = k-2
So the order depends on the original position of the vertex k, e.g for k=0 the positions would be:
1. -3/4
2. 3
3. -1/2
4. 4
5. 7
6. -3
7. -2
So, therefore, the order would be 6 7 1 3 2 4 5
F(x)=3x²-6x+13
a=3, b=-6, c=13
the x coordinate of the vertex is x=-b/(2a), so x=-(-6)/(2*3)=1
when x=1, y=3(1)²-6(1)+13=10
the vertex is at (1,10)
answers are in bold.
Answer:
Find the slope of the line that passes through the points given in the table. The slope is 5.
Use one of the given points to find the y-intercept. Substitute values for x, y, and m into the equation y = mx + b and solve for b. The y-intercept is 1.
Write the formula as a function of n in slope-intercept form. The function is
f(n) = 5n+1 for n in the set of natural numbers.
Step-by-step explanation:
I am not sure what your problem here is.
you understand the inequality signs ?
anyway, to get
6×f(-2) + 3×g(1)
we can calculate every part of the expression separately, and then combine all the results into one final result.
f(-2)
we look at the definition.
into what category is -2 falling ? the one with x<-2, or the one with x>=-2 ?
is -2 < -2 ? no.
is -2 >= -2 ? yes, because -2 = -2. therefore, it is also >= -2.
so, we have to use
1/3 x³
for x = -2 that is
1/3 × (-2)³ = 1/3 × -8 = -8/3
g(1)
again, we look at the definition.
into what category is 1 falling ? the one with x > 2 ? or the one with x <= 1 ?
is 1 > 2 ? no.
is 1 <= 1 ? yes, because 1=1. therefore it is also <= 1.
so we have to use
2×|x - 1| + 3
for x = 1 we get
2×0 + 3 = 3
6×f(-2) = 6 × -8/3 = 2× -8 = -16
3×g(1) = 3× 3 = 9
and so in total we get
6×f(-2) + 3×g(1) = -16 + 9 = -7
