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Brums [2.3K]
3 years ago
10

The graph is supposed to show f(x) = 3 sin (x/4+1) - 1/2. Which of the following are correctly represented in the graph? Select

all that apply.

Mathematics
1 answer:
Ludmilka [50]3 years ago
5 0

Answer:

Options a,b,c are right.

Step-by-step explanation:

Amplitude of the graph = 3 but  shown correctly

Hence option a is right

There is a vertical shift of 1/2 down. Shown correctly

Option b is right

Horizontal shift is 1 unit to the left.  Shown correctly

OPtion c is right

Period is 8 pi  not shown correctly

OPtion d is wrong.

Horizontal expansion is not shown at all for x/4

Hence option e is wrong.


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Solve (x+3)2+(x+3)-2=0
Alex777 [14]

Answer:

pretty sure x=0 if not it might be infinite solutions but im pretty sure its just 0.

I plugged it in and i think it works

Step-by-step explanation:

6 0
3 years ago
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A dealer currently makes a profit of x% of his cost when he sells goods. If the dealer could get his goods for 8% less while kee
CaHeK987 [17]

Answer:

15%

Step-by-step explanation:

Let

Selling price = SP

Cost price = CP

If CP = 100

Profit = selling price - cost price

x% = SP - x% of 100

x% + 100x% = SP

SP = 100 (1 + x/100) (1)

If CP = 92

SP = 92 (1+ {x+10}/100)

Equate both SP

100 (1 + x/100) = 92 (1+ {x+10}/100)

100 + x = 92 + [{23/25} * (x+10)]

{2/25}x = 30/25

x = 30/25 ÷ 2/25

= 30/25 × 25/2

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6 0
3 years ago
PLS HELP ME WITH 12 and the bottom one ASAP!! (SHOW WORK FOR BOTH!!) + LOTS OF POINTS!!!
alexandr1967 [171]
Well, since its isoceles triangle so two equal sides. its either 3 or 7. There is a theorem that states two sides added must be greater than the third. so if the third side is 3. it would be 3+3=6, less than 7, incorrect. therefore the correct side is 7. so the sides are 3, 7, 7. you can add them to confirm the theorem.
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For this case we will define the following:
 A, B: are the values of the triangle angles
 a, b: are the values of the lengths of the sides opposite the angles A, B.
 We then have to use the law of the sine:
 \frac{sinB}{b} = \frac{sinA}{a}
 Clearing the angle B we have:
 sinB = \frac{sinA}{a}*b
 B = arcsin(\frac{sinA}{a}*b)
 Substituting values we have:
 B = arcsin(\frac{sin113}{200}*134)
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