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yawa3891 [41]
3 years ago
8

The zero of F -1(x) in F(x) = x + 3 is

Mathematics
2 answers:
seropon [69]3 years ago
7 0

Answer:

-3

Step-by-step explanation:

Novosadov [1.4K]3 years ago
4 0

We have

y = F(x) = x + 3

so

x = y  - 3

so

F⁻¹(y) = y - 3

which we can write

F⁻¹(x) = x - 3

We solve

0 = F⁻¹(x) = x - 3

x = 3

Answer: 3

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Which of the following is not a typical cost associated with renting?
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3 years ago
Cylinder E has the following dimensions: h = 30 cm and r = 4 cm. Cylinder F has the same radius, but has a height of 5 cm . What
Anarel [89]

Answer:

1759.52cm^3

Step-by-step explanation:

Given data

Cylinder E

h = 30 cm and

r = 4 cm

The expression for the volume is

V= πr^2h

V= 3.142*4^2*30

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Cylinder F

h=5 cm

r = 4 cm

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6 0
2 years ago
According to the scores on the last math test, 80%, or 20, of the students in the class received an A. Find the number of studen
Sladkaya [172]
80% is the same as 4/5, so 4/5 of the class= 20. That means that if there was a picture, 4 parts would be shaded, and 1 would not. 1/4 of 20 is 5, so 1 part= 5 people. 5 x 5 = 25. 25 is the answer. :)
8 0
3 years ago
Read 2 more answers
Given a focus of (4, 5) and directrix of y= -3 , find the equation of the parabola.
andrey2020 [161]
Check the picture below.

notice, the focus point is at 4,5 whilst the directrix line is at y = -3, below the focus point, meaning the parabola is vertical and opening upwards.

keeping in mind that the vertex is "p" distance from either of these fellows, then the vertex is half-way between both of them, notice in the picture, the distance from y = 5 to y = -3 is 8 units, half that is 4 units, thus the vertex 4 units from the focus or 4 units from the directrix, that puts it at (4,1), whilst "p" is 4, since the parabola is opening upwards, is a positive 4 then.

\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}})
\\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------

\bf \begin{cases}
h=4\\
k=1\\
p=4
\end{cases}\implies (x-4)^2=4(4)(y-1)\implies (x-4)^2=16(y-1)
\\\\\\
\cfrac{1}{16}(x-4)^2=y-1\implies \cfrac{1}{16}(x-4)^2+1=y

8 0
3 years ago
Pls help me with this answer
Misha Larkins [42]

student 3 i thin srry if im wrong

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