The height of the doorway is 7 feet less than 5 times its width. if the...
i feel like it’s 8 and 3 but i don’t know if it’s right or if im even doing it right in the first place
height: 8 ft
width: 3 ft
You can check your answer against the problem criteria:
area = (8 ft)(3 ft) = 24 ft^2
height = -7 +5(3) = -7 +15 = 8 . . . . your height
Your numbers match the requirements.
The problem statement asks you to "use polynomials." You have not shown work, so we don't know whether you did. (I like to use a graphing calculator.)
Let x represent the width and y represent the height, both in feet.
y = -7 +5x . . . . . height is 7 ft less than 5 times width
xy = 24 . . . . . . . area is 24 square feet
Using the first equation to substitute for y in the second equation, we have ...
x(-7+5x) = 24
5x^2 -7x -24 = 0 . . . . . put in standard form (the required polynomial)
5x^2 -15x +8x -24 = 0 . . . . . facilitate factoring by grouping*
5x(x -3) +8(x -3) = 0
(5x +8)(x -3) = 0
The positive solution is x = 3. Then y = 24/3 = 8.
The dimensions are 3 ft wide by 8 ft high.
* To factor this, we look for factors of 5·(-24) = -120 that sum to -7. Those factors are -15 and +8. There are several ways to get to the binomial factors once you have these numbers, but the "factor by grouping" method is the most easily explained.
360 different combinations
option a, c,d are correct.
from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have
therefore, by rhs rule,δtzb ≅δszb
by cpctc, sz≅tz
also, from δctz and δasz,
by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz
also,from δasz and δzsb,
by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb
hence, option a, c,d are correct.