There is no option of the box plots, so I have created a version that would represent this data.
To make the box plot you will need the lower extreme, lower quartile, median, upper quartile, and upper extreme.
Please see the attached picture.
Assume r= t not sure if you mistyped this.
t = 15 min
T(t) = 68 (0.5)^(15/10) = 24 Celsius
31 = (0.5)^(t/10) take ln
ln31 = (t/10) ln (0.5) + ln68
t = (ln32 - ln 68/ln0.5) * 10 = 11.3 min
D + 3r = 15
d = r + 3
r + 3 + 3r = 15
4r + 3 = 15
4r = 15 - 3
4r = 12
r = 12/4
r = 3 ....he bought 3 roses, at $ 3 per rose = $ 9 <==
d = r + 3
d = 3 + 3
d = 6....he bought 6 daisies, at $ 1 per daisy = $ 6
Answer:
option C. Angle BTZ Is-congruent-to Angle BUZ
Step-by-step explanation:
Point Z is equidistant from the vertices of triangle T U V
So, ZT = ZU = ZV
When ZT = ZU ∴ ΔZTU is an isosceles triangle ⇒ ∠TUZ=∠UTZ
When ZT = ZV ∴ ΔZTV is an isosceles triangle ⇒ ∠ZTV=∠ZVT
When ZU = ZV ∴ ΔZUV is an isosceles triangle ⇒ ∠ZUV=∠ZVU
From the figure ∠BTZ is the same as ∠UTZ
And ∠BUZ is the same as ∠TUZ
So, the statement that must be true is option C
C.Angle BTZ Is-congruent-to Angle BUZ
So first you have to find the perfect square that matches up with x^2 + 6x
so half of 6, and square it. your perfect square is 9
x^2 + 6x + 9 = 7 + 9
then, condense the left side of the equation into a squared binomial:
(x + 3)^2 = 16
take the square root of both sides:
x + 3 = ± √16
therefore:
x + 3 = ± 4
x = - 3 ± 4
so your solution set is:
x = 1, -7
