What is the answer to this question?
2 answers:
The question:
Mr. Green orders some cookies from a local youth group. Using his order form shown below, complete the last column to find the total to be paid.
Quantity
Description
Unit Price
Total of Line
2
Krispy Kosmos
$2.89
1
Chewy Chocos
$3.19
1
Gooey Globs
$2.99
Total
5.8% tax
Total to be paid
a.
$9.60
b.
$11.96
c.
$12.65
d.
$18.90
Answer:
First, we are going to find the total cost of the groceries before tax:
We know from our table that he bought 2 Krispy Kosmos for $2.89, so the cost of the the 2 Krispy Kosmos was 2($2.89) = $5.78
We also know that he bought only one of the the other two items (Chewy Chocos and Gooey Globs), so to find the total cost of the groceries, we just need to add their respective value to the total value of the 2 Kispy Kosmos:
Total = $3.19+$2.99+$5.78 = $11.96
Now the only thing left is add the sale tax . To do it, we are going to calculate the cost of the tax, and then we'll add that cost to the total.
To calculate the cost of the tax, we need to find 5.8% of $11.96. To do it we are going to convert the tax from percentage to decimal and multiply the resulting decimal by $11.96:
To convert the tax from percentage to decimal, we are going to divide it by 100%: .
Since we have our tax expressed as a decimal, we can find the cost of the tax:
(0.058)($11.96) = $0.69
Now we just need to add the cost of the tax to the total:
Total to be paid = $11.96 + $0.69 = $12.65
We can conclude that the correct answer is c. $12.65
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To find out if a point is inside, on, or outside a circle, we need to substitute the ordered pair into the equation of the circle:
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.
If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.
For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2
A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]
B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]
C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.
If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.
For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2
A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]
B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]
C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]
hope this helps u dear
Please, find attached the image with the complete question
Answer:
Explanation:
The triangle ACB is a right triangle with these features:
- x represents the opposite side to the angle g
- y represents the adjacent side to the angle g
The <em>tangent ratio </em>on an angle in a right triangle is equal to the length of the leg opposite to the angle divided by the length of the leg adjacent to the angle.
Hence:
