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Bingel [31]
3 years ago
15

The graph of an absolute value function opens up and has a vertex of (0, -3).

Mathematics
1 answer:
Anit [1.1K]3 years ago
4 0

Answer:

The domain: all real numbers;

The range:  y\ge -3.

Step-by-step explanation:

The graph of the function y=|x| opens up and has a vertex at (0,0). If the graph of an absolute value function opens up and has a vertex of (0, -3), then the function has an expression

y=|x|-3 (see attached diagram).

The domain of the parent function y=|x| is the set of all real numbers, then the domain of the function y=|x|-3 is the set of all real numbers too.

The range of the parent function y=|x| is y\ge 0, then the range of the function y=|x|-3 is y\ge -3.

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You sell tickets for a fundraiser. You give half of the tickets to your sister to sell. You sell 25 of your half of the tickets
hodyreva [135]

Answer:

84 tickets

Step-by-step explanation:

So first off lets add the tickets we sold to the tickets we have left. Lets x equal the amount of tickets we started with before selling.

25+17=x

Add them up.

42=x

We started with 42 tickets. To find how many tickets we had before spiting the tickets. We need to multiply our tickets by 2 since we had half. Let y be this number

42*2=y

Multiply them up

84=y

We started with 84 tickets.

8 0
3 years ago
What is 2 divided by 9.40
IRINA_888 [86]

Answer:0.212765957

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Please help it’s due tomorrow
denpristay [2]
Answer: 10*(cos(pi) + i*sin(pi))

=======================================

Work Shown:

z1 = 1+i is in the form a+bi where a = 1 and b = 1
r = sqrt(a^2+b^2)
r = sqrt(1^2+1^2)
r = sqrt(2)
theta = arctan(b/a)
theta = arctan(1/1)
theta = pi/4

The polar form for z1 is
z1 = r*(cos(theta) + i*sin(theta))
z1 = sqrt(2)*(cos(pi/4)+i*sin(pi/4)

----------------

z2 = -5+5i is in the form a+bi where a = -5 and b = 5
r = sqrt(a^2+b^2)
r = sqrt((-5)^2+5^2)
r = sqrt(50)
r = 5*sqrt(2)
theta = arctan(b/a)
theta = arctan(5/(-5))
theta = 3pi/4

The polar form for z2 is
z2 = r*(cos(theta) + i*sin(theta))
z2 = 5*sqrt(2)*(cos(3pi/4)+i*sin(3pi/4)
----------------

Now use the rule 
If 
z = a*(cos(b) + i*sin(b)) and w = c*(cos(d)+i*sin(d))
then 
z*w = a*c*(cos(b+d)+i*sin(b+d))

We have
a = sqrt(2)
b = pi/4
c = 5*sqrt(2)
d = 3pi/4

So...

z*w = a*c*(cos(b+d)+i*sin(b+d))
z1*z2 = sqrt(2)*5*sqrt(2)*(cos(pi/4+3pi/4)+i*sin(pi/4+3pi/4))
z1*z2 = 10*(cos(4pi/4)+i*sin(4pi/4))
z1*z2 = 10*(cos(pi)+i*sin(pi))
which is the answer in polar form

7 0
4 years ago
A machine at a chocolate factory wraps 300 chocolate bars in two minutes.
zhuklara [117]
B is basically 1000 divided by 2
3 0
2 years ago
How do you do this it makes no sense
andreev551 [17]

Answer:

Step-by-step explanation:

multiply each, for example, -30 x-15 = 4500

6 0
3 years ago
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