Answer:
A. 
B. 
Step-by-step explanation:
A. 
Where,


Plug in the values to get a polynomial that represents the area of the tool




B. To find area, when x = 4.5 in, plug in the value of x into the equation for the area of the tool.





<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>.</em>
A ratio of 3:2.
The easiest way to go about this is to divide the actual number of wins (18) by its corresponding element in the ratio.
So:

This means every element on the ratio has a value of 6.
The amount of element in the ratio that correspond to losses is 2.
Multiply the actual amount of matches per element in the ratio by the number of elements that represents the losses in the ratio.

Your answer:
<em>"The team lost 12 games."</em>
Answer:
The graph in the attached figure
Step-by-step explanation:
Let
x -----> the number of boys
y -----> the number of girls
----->inequality A
The solution of the inequality A is the shaded area above the solid line
The slope of the solid line is positive
The y-intercept of the solid line is (0,-5)
The x-intercept of the solid line is (5,0)
-----> equation B
The solution of the inequality B is the shaded area above the dashed line
The slope of the dashed line is negative
The y-intercept of the solid line is (0,4)
The x-intercept of the solid line is (4,0)
using a graphing tool
the solution of the system of inequalities is the shaded area between the solid line and the dashed line
see the attached figure
<h3>Answer to Question 1:</h3>
AB= 24cm
BC = 7cm
<B = 90°
AC = ?
<h3>Using Pythagoras theorem :-</h3>
AC^2 = AB^2 + BC ^ 2
AC^2 = 24^2 + 7^2
AC^2 = 576 + 49
AC^2 = √625
AC = 25
<h3>Answer to Question 2 :-</h3>
sin A = 3/4
CosA = ?
TanA = ?
<h3>SinA = Opp. side/Hypotenuse</h3><h3> = 3/4</h3>
(Construct a triangle right angled at B with one side BC of 3cm and hypotenuse AC of 4cm.)
<h3>Using Pythagoras theorem :-</h3>
AC^2 = AB^2 + BC ^ 2
4² = AB² + 3²
16 = AB + 9
AB = √7cm
<h3>CosA = Adjacent side/Hypotenuse</h3>
= AB/AC
= √7/4
<h3>TanA= Opp. side/Adjacent side</h3>
=BC/AB
= 3/√7