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Ghella [55]
2 years ago
12

Which of the following are solutions to the equation below?

Mathematics
1 answer:
larisa86 [58]2 years ago
5 0

Step-by-step explanation:

therefore A is the correct answer

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Which equation represents a quadratic function?<br> y = x2<br> y = 2x<br><br> y =1/x^2
ehidna [41]

Answer:

  y = x²

Step-by-step explanation:

A quadratic function is a polynomial function that has a variable to the power of 2 as the highest degree term. y = 1/x^2 is not a polynomial. y=2x is a linear function.

  y = x^2 is a quadratic function

3 0
3 years ago
In the coordinate plane, you are given JKL with J(2, 3), K(10,4), and L(6, 10). Using the distance formula, classify JKL
Marrrta [24]

Answer:

Perimeter of JKL = 22.55 unit (Approx.)

Step-by-step explanation:

Given:

Coordinate of J = (2,3)

Coordinate of K = (10,4)

Coordinate of L = (6,10)

Find:

Perimeter of JKL

Computation:

Distance = √(x1 - x2)² + (y1 - y2)²

Distance between JK = √(2 - 10)² + (3 - 4)²

Distance between JK = √64 + 1

Distance between JK = √65

Distance between JK = 8.06 unit

Distance between KL = √(10 - 6)² + (4 - 10)²

Distance between KL = √16 + 36

Distance between KL = √52

Distance between KL = 7.21 unit

Distance between LJ = √(6 - 2)² + (10 - 3)²

Distance between LJ = √4 + 49

Distance between LJ = √53

Distance between LJ = 7.28 unit

Perimeter of JKL = Distance between JK + Distance between KL + Distance between LJ

Perimeter of JKL = 8.06 + 7.21 + 7.28

Perimeter of JKL = 22.55 unit (Approx.)

4 0
2 years ago
What's the approximate area of a segment of a circle with a height 6 m and the length of the chord is 20 m? Round your answer to
yuradex [85]
Hello!

The Correct Answer to this would be 100%:

Option "85.4".

(Work Below)

Given:
height = 6m
chord = 20 m

We need to find the radius of the circle.

20 m = 2 √ [ 6m( 2 x radius - 6 m ) ] 
20 m / 2 = 2 √[ 6m( 2 x radius - 6 m ) ] / 2 
10 m = √ [ 6m( 2 x radius - 6 m ) ] 
(10 m)² = √[ 6m( 2 x radius - 6 m ) ] ² 
100 m² = 6 m( 2 x radius - 6 m ) 
100 m² = 12 m x radius - 36 sq m 
100 m² + 36 m² = 12 m x radius - 36 m² + 36 m² 
136 m² = 12 m x radius 
136 m² / 12 m = 12 m x radius / 12 m 
<span>11.333 m = radius 
</span>
the area beneath an arc: 

Area = r² x arc cosine [ ( r - h ) / r ] - ( r - h ) x √( 2 x r x h - h²<span> ). 
</span>
r² = (11.333 m)² = 128.444 m² 
r - h= 11.333 m - 6 m = 5.333 m 
r * h = 11.333 m x 6 m = 68 m²

Area = 128.444 m² x arc cosine [ 5.333 m / 11.333 m ] - 5.333 m x √[ 2 x 68 m² - 36 m² ] 

Area = 128.444 m² x arc cosine [ 0.4706 ] - 5.333 m x √ [ 100m² ] 

Area = 128.444 m² x 1.0808 radians - 5.333 m x 10 m 

Area = 138.828 m² - 53.333 m² 

Area = 85.4 m<span>²
</span>

Hope this Helps! Have A Wonderful Day! :)


And as Always...

7 0
3 years ago
Which is the equation of the graphed line written in standard form? y = x x – 1/2y = 0 x – y = 0 y =1/2x
AysviL [449]

The standard for of this equation is x – y = 0

Firstly, options A and D are not in standard form, so they are not options.

To decided between B and C, we choose a point on the line and see if it works. For example, we'll use (2, 2).

x – y = 0

2 - 2 = 0

0 = 0

This is true for option C. Therefore, it is correct.

5 0
3 years ago
Compute the total and annual return on the following investment Two years after paying 53500 for shares in a startup company you
wel

Answer:

-46.91%

Step-by-step explanation:

Given:

Cost of buying the shares = $53,500

Selling cost of the shares = $3,300

Number of years = 2

Since the selling cost is less than the buying cost, therefore the statement for the loss can be verified

Now,

The loss = Selling cost - cost of buying = $3,300 - $53,500 = - $50,200

thus, the loss per year = \frac{\textup{50,200}}{\textup{2}} = -$25100

Hence,

Rate of return = \frac{\textup{Loss per year}}{\textup{Cost of buying}}\times100

or

Rate of return = \frac{\textup{-25100}}{\textup{53,500}}\times100

or

Rate of return = -46.91%

4 0
3 years ago
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