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LuckyWell [14K]
3 years ago
9

15 points ! precalc question with graph

Mathematics
2 answers:
Umnica [9.8K]3 years ago
7 0

By definition, we have to:

1) the cut points with the horizontal axis are given by pairs ordered in the following way:

(x, 0)

2) the cutting points with the vertical axis are given by pairs ordered in the following way:

(0, y)

Therefore, using the definition we have:

Cut with the horizontal axis:

(- 1, 0)

(3, 0)

Cut with vertical axis:

(0, -3)

Answer:

option 2

Semmy [17]3 years ago
6 0

x-intercepts are where the graph crosses the x-axis. These are (-1, 0) and (3, 0) from the graph.

y-intercepts are where the graph crosses the y-axis. This is (0, -3)  from the graph.

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FUNDRAISING A school is raising money by selling calendars for $20 each. Mrs. Hawkins promised a party to whichever of her Engli
____ [38]

Rj, this is the solution to question 19:

• 1st period class sold 60 calendars in total

,

• 2nd period class sold 123 calendars in total

We calculate the average per week, this way:

• 1st period class sold 60/4 = 15 calendars per week

,

• 2nd period class sold 123/4 = 30.75 calendars per week

In consequence the multiplication equation that represents this situation is:

15w + 30.75w = 183, where w is the number of weeks Mrs. Hawkins' classes sold calendars.

This is the solution to question 20:

We already know the average number of calendars that 1st and 2nd period classes sold per week. Now, let's calculate the average for 3nd 4th peiod classes, this way:

• 3rd period class sold 89 calendars in total

,

• 4th period class sold 126 calendars in total

Therefore, the average is:

• 3rd period class sold 89/4 = 22.25 calendars per week

,

• 4th period class sold 126/4 = 31.5 calendars per week

Summarizing, we have:

15 + 30.75 + 22.25 + 31.5 = 99.5 calendars was the average number sold by all her classes in a week.

•

7 0
1 year ago
Help, please in this problem
Nikolay [14]

Answer:

x=9.768

y=6.972

Step-by-step explanation:

For this problem we have to use the trig relationships of cos and sin to figure out the lengths. Cos is equal to adjacent/hypotenuse so we can set it as x/r=.814 and since r is equal to 12 we can do 12 times .814 to get x.

We do a similar process for sin but sin is equal to opposite/hypotenuse so we can set up the equation y/r=.581 and we simply multiply both sides by 12 to get 12*.581 to get y.

Also for future reference adjacent and hypotenuse are based on the angle at use, since ∅ is on the bottom left x is the adjacent side and y is the opposite side.

3 0
3 years ago
Which of the following is equivalent to tan2θcos(2θ) for all values of θ for which tan2θcos(2θ) is defined?
Aloiza [94]

Answer:

2sin²θ - tan²θ

Step-by-step explanation:

Given

tan²θcos(2θ)

Required

Simplify

We start by simplifying cos(2θ)

cos(2θ) = cos(θ+θ)

From Cosine formula

cos(A+A) = cosAcosA - sinAsinA

cos(A+A) = cos²A - sin²A

By comparison

cos(2θ) = cos(θ+θ)

cos(2θ) = cos²θ - sin²θ ----- equation 1

Recall that cos²θ + sin²θ = 1

Make sin²θ the subject of formula

sin²θ = 1 - cos²θ

Substitute sin²θ = 1 - cos²θ in equation 1

cos(2θ) = cos²θ - (1 - cos²θ)

cos(2θ) = cos²θ - 1 +cos²θ

cos(2θ) = cos²θ + cos²θ - 1

cos(2θ) = 2cos²θ - 1

Substitute 2cos²θ - 1 for cos(2θ) in the given question

tan²θcos(2θ) becomes

tan²θ(2cos²θ - 1)

Open brackets

2cos²θtan²θ - tan²θ

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

Simplify tan²θ

tan²θ = (tanθ)²

Recall that tanθ =  sinθ/cosθ

So, we have

tan²θ = (sinθ/cosθ)²

tan²θ = sin²θ/cos²θ

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

Substitute sin²θ/cos²θ for tan²θ

2cos²θtan²θ - tan²θ becomes

2cos²θ(sin²θ/cos²θ) - tan²θ

Open bracket (cos²θ will cancel out cos²θ) to give

2(sin²θ) - tan²θ

2sin²θ - tan²θ

Hence, the simplification of tan²θcos(2θ) is 2sin²θ - tan²θ

Option E is correct

7 0
3 years ago
Question 28
alexandr402 [8]

Answer:

Given Rate of interest is r=8%=0.08

Principal Amount is A=5,000

Time is t years

Interest is compounded yearly twice ⟹n=2

Amount =P(1+

n

r

)

nt

=5000×(1+

2

0.08

)

2t

=5,408

(1.0816)

t

=

5000

5408

⟹t=1

7 0
3 years ago
The point (–3, –5) is on the graph of a function. What equation must be true regarding the function?
Komok [63]

The equation that must be true regarding the function is a. f(–3) = –5

<h3>How to explain the information?</h3>

The point (–3, –5) is on the graph of a function. Which equation must be true regarding the function?

a. f(–3) = –5

b. f(–3, –5) = –8

c. f(–5) = –3

d. f(–5, –3) = –2

The question is what does the point (-3, -5) correspond to on the graph of the function.

If we have a point on a graph in the Cartesian coordinate system then that point consists of coordinates (x, y). In other words, y=f(x) and x so (x, f(x)) where x is a x-coordinate and y=f(x) is y-coordinate.

Hence if we have a point (-3, -5) the corresponding coordinates are x=-3 and y=f(x)=-5.

Therefore the correct answer is f(-3)=-5.

Learn more about equation  on:

brainly.com/question/2972832

#SPJ1

7 0
2 years ago
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