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Assoli18 [71]
3 years ago
13

Graph the following using the parabola toolf(x) = -(x+3)^2+5​

Mathematics
1 answer:
Allisa [31]3 years ago
7 0
The Max point is at (-3,5)

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How many 9's are in 99
MrRissso [65]

Answer:

11 or 2

Step-by-step explanation:

11: You take 99/9 to see how many times 9 could go into 99, which would be 11.

2: You just look at the number and you see that there are 2 9s in 99.

-I hope this helps!-

-I didn't know which answer you were looking for so I put both!-

-Please mark as brainliest!-

-Good luck!-

4 0
3 years ago
Last year, sales at a book store increased from $5,000 to $10,000. This year, sales decreased to $5,000 from $10,000. What perce
suter [353]

Sales increased 50% last year, from $5,000 to $10,000. When sales dropped from $10,000 to $5,000 this year, sales decreased 50%

5 0
3 years ago
a news paper carrier has 10.30 in change. he has two more quarters than dimes but three times as many nickles as quarters. how m
Karolina [17]
A) .05n + .10d + .25 q = 10.30
B) d +2 = q
C) n = 3q
Substituting equations B and C into A
A) .05*3q + .10(q-2) + .25 q = 10.30
A) .15q + .1q -.2 + .25q = 10.30
A) .50q = 10.50
21 quarters

C) n = 3q
C) n = 3*21
63 nickels

B) d +2 = q
d + 2 = 21
19 dimes


4 0
3 years ago
How do you factor 7+14
mote1985 [20]

You could factor a 7 out of 7 + 14 to get 7(1 + 2).

This would equal the same thing as 7 + 14, which is 21.

7 * 3 = 21.

21 can be factored like so:

1 * 21 = 21

3 * 7 = 21


5 0
4 years ago
What is the range of the function y = x 2?
Leno4ka [110]

Answer:

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

Step-by-step explanation:

Given the function

y=x^2

  • We know that the range of a function is the set of values of the dependent variable for which a function is defined.

\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)

\mathrm{If}\:a

\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v

a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)

f\left(x\right)\ge \:0

Thus,

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

5 0
3 years ago
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