Answer:
(-5,24)
Step-by-step explanation:
2f+3g-h
first we substitute in the numbers
2(-3)+3(1)-2
-6+3-2
-3-2
-5
Then again
2(5)+3(4)-(-2)
10+12+2
22+2
24
hope this helps!:)
Answer:500
Step-by-step explanation:26,000/52=500 i did<em> $26,000</em> divided by 52 because there are<em> </em><em>52 weeks in a year</em> and it gave me <em>500</em> so that is how i got the answer
Answer:
The correct answer is x = 17.
Step-by-step explanation:
If EF bisects DEG, this means that angles DEF and angles FEG are congruent, and they each make up half of angle DEG.
Therefore, we can set up the equation:
DEF + FEG = DEG
However, since we know that DEF and FEG represent the same value, we can change this equation into the following:
2(DEF) = DEG
Now, we can substitute in the expressions that we are given:
2(3x+1) = 5x + 19
To simplify, we should first use the distributive property on the left side of the equation.
6x + 2 = 5x + 19
Our next step is to subtract 5x from both sides of the equation.
x + 2 = 19
Finally, we can subtract 2 from both sides of the equation to get x by itself on the left side.
x = 17
Therefore, the value of x is 17.
Hope this helps!
The equation of a circle whose centre (
) and radius 
is 
What is equation of a circle?
A circle is a closed curve drawn from a fixed point called centre of the circle. The distance between the centre of the circle and the arc of the circle is called the radius of the circle.
The equation of a circle with centre 
radius 
is
Given center = 
and Radius =
Put the Given value in the equation:
= 
= 
= 
= 
= 
= 
So, the equation of the circle whose centre is and Radius is
To read more about the equation of a circle. you can follow:
brainly.com/question/351704
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Answer:
Step-by-step explanation:
The width of rectangle is the diameter of the semi-circle part
Area of one semicircle is given by 
Total area of semi circle will be 
Substituting 74 m for d and 
as 3.14 we obtain
Total area semi-circle=
Area of rectangle is given by the product of length and width
Rectangular area=
Total area of rectangular and semi-circles will be
Therefore, area of training field is
