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Ksenya-84 [330]
3 years ago
6

PLEASEEEEE HELPPPP ASPPPPPPP

Mathematics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

The answer is A because a number on it's own is a constant (30) and a number with a variable is a coefficient (12h)

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The side of one square is equal to 3m and its diagonal is equal to the side of a second square. Find the diagonal of the second
navik [9.2K]

Answer:

6m

Step-by-step explanation:

the side of first square = 3 m

area of first square = 3^2= 9 m^2

let the side of second square = x m

since diagonal of first square = x m as per the question

in a square each angle is 90 degree

therefore by applying pythagorus theorem ,

diagonal ^2 = side^2 + side ^2

diagonal^2 = 3^2+3^2=9+9

DIAGONAL= square root of 18

diagonal= 3\sqrt{2} m = side of second square

therefore in second square ,

one angle is 90  degree , therefore by applying pythagorue theorem

diaqonal^2 = side ^2 + side ^2 = (3\sqrt{2})^2 + ( 3\sqrt{2})^2

diagonal ^2 = 18 + 18

diagonal = square root of 36

therefore diagonal of second square is 6m

hope it helps you... please mark me as the brainliest..

8 0
3 years ago
Read 2 more answers
Factor the GCF from this polynomial <br><br> 40x-10
kipiarov [429]

10(4x-1) is the answer to your question my friend

8 0
3 years ago
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The Cartesian coordinates of a point are given. (a) (−5, 5) (i) Find polar coordinates (r, θ) of the point, where r &gt; 0 and 0
Alex73 [517]

Answer:

a)

(i) The coordinates of the point in polar form is (5√2 , 7π/4)

(ii) The coordinates of the point in polar form is (-5√2 , 3π/4)

b)

(i) The coordinates of the point in polar form is (6 , π/3)

(ii) The coordinates of the point in polar form is (-6 , 4π/3)

Step-by-step explanation:

* Lets study the meaning of polar form

- To convert from Cartesian Coordinates (x,y) to Polar Coordinates (r,θ):

1. r = √( x2 + y2 )

2. θ = tan^-1 (y/x)

- In Cartesian coordinates there is exactly one set of coordinates for any

 given point

- In polar coordinates there is literally an infinite number of coordinates

 for a given point

- Example:

- The following four points are all coordinates for the same point.

# (5 , π/3) ⇒ 1st quadrant

# (5 , −5π/3) ⇒ 4th quadrant

# (−5 , 4π/3) ⇒ 3rd quadrant

# (−5 , −2π/3) ⇒ 2nd quadrant

- So we can find the points in polar form by using these rules:

 [r , θ + 2πn] , [−r , θ + (2n + 1) π] , where n is any integer

 (more than 1 turn)

* Lets solve the problem

(a)

∵ The point in the Cartesian plane is (-5 , 5)

∵ r = √x² + y²

∴ r = √[(5)² + (-5)²] = √[25 + 25] = √50 = ±5√2

∵ Ф = tan^-1 (y/x)

∴ Ф = tan^-1 (5/-5) = tan^-1 (-1)

- Tan is negative in the second and fourth quadrant

∵ 0 ≤ Ф < 2π

∴ Ф = 2π - tan^-1(1) ⇒ in fourth quadrant r > 0

∴ Ф = 2π - π/4 = 7π/4

OR

∴ Ф = π - tan^-1(1) ⇒ in second quadrant r < 0

∴ Ф = π - π/4 = 3π/4

(i) ∵ r > 0

∴ r = 5√2

∴ Ф = 7π/4 ⇒ 4th quadrant

∴ The coordinates of the point in polar form is (5√2 , 7π/4)

(ii) r < 0

∴ r = -5√2

∵ Ф = 3π/4 ⇒ 2nd quadrant

∴ The coordinates of the point in polar form is (-5√2 , 3π/4)

(b)

∵ The point in the Cartesian plane is (3 , 3√3)

∵ r = √x² + y²

∴ r = √[(3)² + (3√3)²] = √[9 + 27] = √36 = ±6

∵ Ф = tan^-1 (y/x)

∴ Ф = tan^-1 (3√3/3) = tan^-1 (√3)

- Tan is positive in the first and third quadrant

∵ 0 ≤ Ф < 2π

∴ Ф = tan^-1 (√3) ⇒ in first quadrant r > 0

∴ Ф = π/3

OR

∴ Ф = π + tan^-1 (√3) ⇒ in third quadrant r < 0

∴ Ф = π + π/3 = 4π/3

(i) ∵ r > 0

∴ r = 6

∴ Ф = π/3 ⇒ 1st quadrant

∴ The coordinates of the point in polar form is (6 , π/3)

(ii) r < 0

∴ r = -6

∵ Ф = 4π/3 ⇒ 3rd quadrant

∴ The coordinates of the point in polar form is (-6 , 4π/3)

6 0
3 years ago
Please help me out with this please
ki77a [65]

Answer:

P = 9 is the max value

Step-by-step explanation:

Sketch

2x + 4y = 10

with x- intercept = (5, 0) and y- intercept (0, 2.5)

x + 9y = 12

with x- intercept = (12, 0) and y- intercept = (0, \frac{4}{3} )

Solve

2x + 4y = 10 and x + 9y = 12 to find the point of intersection at (3, 1)

The region corresponding to the solution of the system of constraints

Has vertices at (0, \frac{4}{3}), (0, 0) , (5, 0) and (3, 1)

Now evaluate the objective function at each vertex.

(0, 0) can be excluded as it will not give a maximum

(5, 0) → P = 5 + 0 = 5

(0, \frac{4}{3}) → 0 + 8 = 8

(3, 1) → 3 + 6(1) = 3 + 6 = 9 ← maximum value

Thus the maximum value is 9 when x = 3 and y = 1

7 0
3 years ago
What is the answer to 3/10x120
Elza [17]

Answer:

3/10 * 120

= 3 * 120/10

= 3 * 12

= 36

4 0
3 years ago
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