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ANEK [815]
3 years ago
15

WILL MARK BRAINLIEST IF CORRECT

Mathematics
1 answer:
RideAnS [48]3 years ago
7 0
The answer would be “D” because it’s is 4 less than 7j.
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Mackenzie spent a total of $17.50 on Saturday afternoon.She bought a movie ticket for $7.25 and snacks for $4.95.She spent the r
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17.50-7.25-4.95=5.30/2=2.65
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Lynn multiplies 7/8 Times a number the product is less than 7/8 which could be Lynn's number
Brums [2.3K]

Answer: 6/8 <;3..?

Step-by-step explanation:

7 0
3 years ago
Identify the domain and range of each of the graphs below.<br><br><br> Domain: <br><br><br> Range:
Anna11 [10]

Answer:

Domain → (-∞, ∞)

Range → (-∞, ∞)

Step-by-step explanation:

Domain of a function is defined by the x-values or input values of the graph.

Similarly, Range of the function is defined by the y-values or output values from the graph.

From the picture attached,

x - values for the line shown in the graph vary from negative infinity to positive infinity.

Therefore, Domain of the function will be → (-∞, ∞)

And for every x-value there is a y-value, so Range of the function will be same as domain (varying from negative infinity to positive infinity) → (-∞, ∞)

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3 years ago
tanmayi wants to raise $175 for school fundraisder.she has raised $120 so far.How much more does she need to reach her goal?
Bezzdna [24]
175>120+x

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3 years ago
Match each of the parametric equations 1-5 with the curve it represents from the list A through E below. 1. r=3 2. r=2sinθ 3. θ=
ira [324]

Answer:

Step-by-step explanation:

1. r=3

To convert from polar coordinates (r,θ) to rectangular coordinates (x,y), we use the following equations

x=rCosθ

Cosθ=x/r

y=rSinθ

Sinθ=y/r

Also, r²=x²+y²

When r =3,

Then we have

r=3,

x²+y²=3

All points with r = 3 are at

distance 3 from the origin, so r = 3 describes the circle of radius 3, with center at the origin (0,0).

Option C

2. r=2sinθ

When r=2sinθ

To convert from polar coordinates (r,θ) to rectangular coordinates (x,y), we use the following equations

x=rCosθ

Cosθ=x/r

y=rSinθ

Sinθ=y/r

Also, r²=x²+y²

Now, apply the given information

r=2sinθ

Since Sinθ=y/r

r=2y/r

Cross multiply

r²=2y

x²+y²=2y

x²+y²-2y =0

x²+ (y-1)² -1 =0

x²+(y-1)²=1²

Then,

It is a circle with center (0,1) and radius 1.

Because the sine is periodic, we know that we will get the entire curve for values of θ in [0, 2π). As θ runs from 0 to π/2, r increases

from 0 to 2. Then as θ continues to π, r decreases again to 0. When θ runs from π to

2π, r is negative, and it is not hard to see that the first part of the curve is simply traced

out again, so in fact we get the whole curve for values of θ in [0, π). Thus, check attachment for curve, Now, this suggests that the curve could possibly be a circle,

and if it is, it would have to be the circle x² + (y − 1)² = 1. Having made this guess, we can easily check it. First we substitute for x and y to get (r cos θ)² + (r sin θ − 1)² = 1;

expanding and simplifying does indeed turn this into r = 2 sin θ.

Option E

A circle of radius 1 with center on the x-axis

3. θ=π/4

This is a point on the circle

p=(r, θ)

So, r=0

x = rCosπ/4=rCos45

x=r√2 /2

and

y = rsin(π/4) = rSin45

y=r√2 /2

This makes it very

easy to convert equations from rectangular to polar coordinates.

x²+y²= (r√2/2)²+(r√2/2)²

x²+y²= 2r²/4 + 2r²/4

x²+y²= r²

Since r=0

x²+y²= 0

x²=y²

Therefore, x=y

Option D

The straight line y=x

4. r=4θ

Here the distance from the origin

exactly matches the angle, so a bit of thought makes it clear that when θ ≥ 0 we get the spiral of Archimedes ( check attachment ) When θ < 0, r is also negative, and so the full graph is the right hand picture in the figure. Check attachment

The correct option is B

None of the above

5. r=2Cosθ

To convert from polar coordinates (r,θ) to rectangular coordinates (x,y), we use the following equations

x=rCosθ

Cosθ=x/r

y=rSinθ

Sinθ=y/r

Also, r²=x²+y²

Now, apply the given information

r=2cosθ

Since cosθ=x/r

r=2x/r

Cross multiply

r²=2x

x²+y²=2x

x²+y²-2x=0

(x-1)² + y² -1 =0

(x-1)² + y² = 1²

Then,

It is a circle with center (1,0) and radius 1.

A circle of radius 1 with center on the y-axis

Option A is correct

7 0
3 years ago
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