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Wittaler [7]
2 years ago
7

PLEASE HELPPPP(BRAINLIEST FOR RIGHT ANSWER)

Mathematics
2 answers:
Thepotemich [5.8K]2 years ago
8 0

Answer:

The relation between polar and Cartesian coordinates are

r2=x2+y2

tan

θ

=

y

x

,

x

=

r

cos

θ

,

y

=

r

sin

θ

r

=

8

sin

θ

∴

r

⋅

r

=

r

⋅

8

sin

θ

or

r

2

=

8

r

sin

θ

∴

x

2

+

y

2

=

8

y

or

x

2

+

y

2

−

8

y

=

0

or

x

2

+

y

2

−

8

y

+

16

=

16

or

x

2

+

(

y

−

4

)

2

=

4

2

Rectangular form is

x2+(y-4)2=42

graph{x^2+(y-4)^2=16 [-20, 20, -10, 10

Step-by-step explanation:

i created it that way so yall will take a longer time to answer

Fynjy0 [20]2 years ago
4 0
The last one is the answer
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For what value of k does the equation 6(x + 1) + 2 = 3(k5x + 1) + 3 have no solution?
NARA [144]

Answer:

k = (6/15)

Step-by-step explanation:

The equation is:

6*(x + 1) + 2 = 3*(k*5*x + 1) + 3

To have no solutions, we need to have something like:

x + 7 = x + 4

where we can remove x in both sides and end with

7 = 4

So this equation is false, meaning that there is no value of x such that this equation is true, then the equation has no solutions.

First, let's try to simplify our equation:

6*(x + 1) + 2 = 3*(k*5*x + 1) + 3

6*x + 6 + 2 = 3*k*5*x + 3*1 + 3

6*x + 8 = 15*k*x + 6

if 15*k = 6, then the system clerly has no solution.

then:

k = 6/15

then we get:

6*x + 8 = (6/15)*15*x + 6

6*x + 8 = 6*x + 6

8 = 6

The system has no solutions.

8 0
2 years ago
triangle ABC is reflected across the y-axis and then dilated by a factor of 1/2 centered at the origin. which statement correctl
Nadya [2.5K]

Answer:

Step-by-step explanation:

The reflection preserves the side lengths and angles of triangle ABC. The dilation preserves angles but not side lengths.

5 0
2 years ago
An engineer is designing a large steel pad to be installed on the deck of an aircraft carrier. Its total volume will be 18 yd^3.
Shtirlitz [24]

Answer:

The cost of the material for the full-sized pad is;

d) 222,750.00

Step-by-step explanation:

The parameters of the steel pad and the scale model are;

The volume of the deck = 18 yd³

The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick

The weight of the sample, m₂ = 75 lbs

The cost of the steel, P = $55/lb

The dimensions of the sample in yards are;

6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0.\overline 1 yards

The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0.\overline 1 yd. =  (1/3) yd.³

The density of the sample material, ρ = m₂/V₂

∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³

Therefore, the density of the material of the pad, ρ = 225 lbs/yd³

The mass of the steel  pad, m₁ = ρ × V₁

∴ The mass of the steel  pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs

The cost of the material for the full-sized steel pad, C = m₁ × P

∴ C = 4,050 lbs × $55/lb = $222,750

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