Answer:
d. 1.5%
Step-by-step explanation:
Given - The track and field coach records Ian's 400-meter race times during three practices.
Practice 1 2 3
Time (s) 63 60 58
To find - How much greater was the percent of change from Practice 1 to Practice 2 than from Practice 2 to Practice 3? Round to the nearest tenth.
a.
7.9%
b. 4.8%
c. 3.3%
d. 1.5%
Proof -
% change from Practice 1 to Practice 2 change = %
= % = - 4.761904762 %
⇒% change from Practice 1 to Practice 2 change = - 4.762 %
Now,
% change from Practice 2 to Practice 3 change = %
= % = - 3.334 %
⇒% change from Practice 2 to Practice 3 change = - 3.3334 %
So,
- 4.461904762% - (- 3.3334%) = -4.461904762 + 3.3334 % = -1.4285 % ≈ 1.5%
∴ we get
Value = 1.5%
The correct option is - d. 1.5%
Answer:
1638m²
Step-by-step explanation:
42×39= 1628m²
AC is a tangent so by definition, it touches the circle at exactly one point (point C) and forms a right angle at the tangency point. So angle ACO is 90 degrees
The remaining angle OAC must be 45 degrees because we need to have all three angles add to 180
45+45+90 = 90+90 = 180
Alternatively you can solve algebraically like so
(angle OAC) + (angle OCA) + (angle COA) = 180
(angle OAC) + (90 degrees) + (45 degrees) = 180
(angle OAC) + 90+45 = 180
(angle OAC) + 135 = 180
(angle OAC) + 135 - 135 = 180 - 135
angle OAC = 45 degrees
Side Note: Triangle OCA is an isosceles right triangle. It is of the template 45-45-90.
The answer is division because u divide 3 from both side of the equation so that make the answer B
.25p + 5 = .50p + 2
-.25p = -3
-p = -12
p= 12 problems