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Ronch [10]
3 years ago
11

The red function was transformed into the blue function. Which transformations have occurred?

Mathematics
1 answer:
marin [14]3 years ago
5 0

Answer:

Translated 3 units down 2 units left.

Step-by-step explanation:

It looks like the functions are the same, so if you see, it was translated 3 units down and 2 units left.

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Label 1/4 on the number line. Label 1/2 on the number line .
KIM [24]

Answer:

it goes 0 then 1/4 in the first notch 1/2 in the second notch 3/4 in the third notch then the number 1 on the fourth

Step-by-step explanation:

Hope that helps.

5 0
3 years ago
What number must you multiply both sides of 3/4a=24 by to get equivalent equation a=32??
Mamont248 [21]
4 because 4 times 24 equals 96 then 96 divided by 3 equals 32
3 0
3 years ago
Applications 24. A rocket is launched into the air. Its height in feet, after x seconds, is given by the equation The starting h
SVEN [57.7K]

The given function is

h(x)=-16x^2+300x+20

According to this function, the starting height of the rocket is 20 feet because that's the initial condition of the problem stated by the independent term.

Additionally, we find the maximum height by calculating the vertex of the function V(h,k).

h=-\frac{b}{2a}

Where a = -16 and b = 300.

\begin{gathered} h=-\frac{300}{2(-16)}=\frac{300}{32}=\frac{150}{16}=\frac{75}{8} \\ h=9.375 \end{gathered}

Then, we find k by evaluating the function

\begin{gathered} k=-16(9.375)^2+300(9.375)+20 \\ k=-1406.25+2812.5+20=1426.25 \end{gathered}

Hence, the maximum height is 1426.25 feet.

At last, to know the time need to hit the ground, we just use h=9.375 and we multiply it by 2

t=2\cdot9.375=18.75

Hence, the rocket hits the ground after 18.75 seconds.

6 0
1 year ago
Solve the following quadratic equation for all values of x in simplest form.<br> 2(x – 5)^2 = 32
Vadim26 [7]
x1=1 and x2=9 Whenever you see x put the two answers there and it’s equal to 32.... I hope it helped???? :)
6 0
3 years ago
hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams a
nalin [4]

Answer:

a. 205320

b. 34220

c. 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

Step-by-step explanation:

a) The number of ways to dustribute exams among the TA's is:

n / (n - r)!

n= number of things to choose from

r= Choosing r number

60P3= 60! / (60 - 3)!

(60)(59)(58)(57)! / (57)!

=205320

B) The number of ways to dustribute the exams among the TA's is:

n! /(n - r)! r!

60C3= 60! /(60 - 3)! 3!

= 60!/ 57! 3!

= 60 × 59 × 58 / 3 × 2 × 1

= 34220

C) The required number of ways is:

60C25 + 60C20 + 60C15

= 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

6 0
3 years ago
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