The equations that can be used are 10T + 5S = 190 and T + S = 30.
<h3><u>Equations</u></h3>
Given that the girls tennis team was interested in raising funds for an upcoming trip, and the team sold tumblers for $10 and sun hats for $5, and when the sales were over, the team had earned $190 and sold 30 total products, which included a mix of tumblers and hats, to determine which equations can be used to represent the situation, the following calculations must be made:
- T + S =190
- -It cannot be used because it has any relationship with the price of the products.
- 10T + 5S = 30
- -It cannot be used because it only considers the quantity variable.
- T + S = 30
- -It can be used as it shows the amount of products sold.
- 10T + 5S = 190
- -It can be used because it relates the total price to the quantity of each product.
- T + S = 15
- -It cannot be used because it only considers the price variable.
- 5T + 10S = 190
- -It cannot be used because it erroneously relates the price of each product.
Therefore, the equations that can be used are 10T + 5S = 190 and T + S = 30.
Learn more about equations at brainly.com/question/26511270.
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Answer:
3a by 4a
Step-by-step explanation:
For dimensions L and W, the area and perimeter are ...
A = LW = 12a^2
P = 2(L+W) = 14a
Using the second equation, we can find L:
L +W = 7a . . . . . divide by 2
L = 7a -W
Substituting into the area formula gives the quadratic ...
(7a -W)(W) = 12a^2
W^2 -7aW +12a^2 = 0 . . . . arrange in standard form
(W -3a)(W -4a) = 0 . . . . . . . factor (find factors of 12 that total 7)
Then we have the two solutions ...
W = 3a, L = 4a
W = 4a, L = 3a
The rectangle dimensions are 3a by 4a.
For fog or f(g(x)), plug g(x) into f(x):
2(3x+2)²+x
Then simplify:
2( (3x+2)(3x+2) )+x
2( 9x²+12x+4 )+x
18x²+24x+8 +x
18x²+25x+8
Then you can simplify this further using factoring methods to:
(2x+1)(9x+8)
Hope this helps!! :)
Answer:
1) difference of squares, then its factors are (x+4)(x+2)
2) 3 (x+4)/2 (x-4)(x+4)
3/2 (x-4)
3) 3x+11/5x-9
Check the picture below.
now, we know that the slanted legs are congruent, since it's an isosceles trapezoid, we also know that the bases are the parallel sides, so, the "altitude" or distance from those bases are the same length, for each of those triangles in the picture.
now, the bases are parallel, that means the altitude segment is perpendicular to the base, the longest side at the bottom, so, we end up with a right-triangle that has a Hypotenuse and a Leg, equal to the other triangle's.
thus, by the HL theorem for right triangles, both of those triangles are congruent, and if the triangles are congruent, all their sides are also, including the ones on the base.