78×.33=25.74 busses are new
25.74+15=40.74 new busses
78+15=93 total busses
40.74/78=0.5223076923= 52% will be new with the new 15 busses
Answer:
x = -14
Step-by-step explanation:
x + 6 = 4x/7
7x + 42 = 4x
7x - 4x = -42
3x = -42
x = -42 ÷ 3
x = -14
Take half of the middle coefficient.
Then square it.
-12/2 = -6
(-6)^2 = 36
c = 36
QUESTION 3
The sum of the interior angles of a kite is
.
.
.
.
.
But the two remaining opposite angles of the kite are congruent.

.
.
.
.
QUESTION 4
RH is the hypotenuse of the right triangle formed by the triangle with side lengths, RH,12, and 20.
Using the Pythagoras Theorem, we obtain;





QUESTION 5
The given figure is an isosceles trapezium.
The base angles of an isosceles trapezium are equal.
Therefore
QUESTION 6
The measure of angle Y and Z are supplementary angles.
The two angles form a pair of co-interior angles of the trapezium.
This implies that;



QUESTION 7
The sum of the interior angles of a kite is
.
.
.
.
.
But the two remaining opposite angles are congruent.

.
.
.
.
QUESTION 8
The diagonals of the kite meet at right angles.
The length of BC can also be found using Pythagoras Theorem;




QUESTION 9.
The sum of the interior angles of a trapezium is
.
.
.
But the measure of angle M and K are congruent.
.
.
.
.
Answer:
-4, -6, -3, -5, -1. The inequality solved for n is n ≥ -6.
Step-by-step explanation:
Substitute all the values in the equation.
n/2 ≥ -3
-10/2 ≥ -3
-5 is not ≥ -3.
n/2 ≥ -3
-7/2 ≥ -3
-3.5 is not ≥ -3.
n/2 ≥ -3
-4/2 ≥ -3
-2 is ≥ -3.
n/2 ≥ -3
-9/2 ≥ -3
-4.5 is not ≥ -3.
n/2 ≥ -3
-6/2 ≥ -3
-3 is ≥ -3.
n/2 ≥ -3
-3/2 ≥ -3
-1.5 is ≥ -3.
n/2 ≥ -3
-8/2 ≥ -3
-4 is not ≥ -3.
n/2 ≥ -3
-5/2 ≥ -3
-2.5 is ≥ -3.
n/2 ≥ -3
-2/2 ≥ -3
-1 is ≥ -3.
To solve the inequality n/2 ≥ -3 for n, do these steps.
n/2 ≥ -3
Multiply by 2.
n ≥ -6.