Translate The phrase into an algebraic expression.

In a school contest, each student draws a number from 1 to 5 out of a large basket. The outcomes are shown in the table below.

12345 [234]

Answer:

The answer is 88/407

Step-by-step explanation:

Their are 5 possible outcomes 1/83, 2/79, 3/88, 4/72, and 5/85. Trust me its 88/407

4 0

2 years ago

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What expression is equivalent to (4-x)+3

sattari [20]

The answer would be B

5 0

3 years ago

It costs $5.15 to buy 1/4 oz of walnuts. how much would it cost to purchase 4.5 oz of walnuts

nalin [4]

Answer:

Step-by-step explanation: 1/4 is 0.25 so 0.25 X 18 = 4.5 so then 18X 5.15 = 81

3 0

3 years ago

What is the equation of a parabola with a directrix of y=2 and a focus point of 0,-2

KiRa [710]

Hope this helped. :)

Any point, (x0,y0)(x0,y0) on the parabola satisfies the definition of parabola, so there are two distances to calculate:

Distance between the point on the parabola to the focusDistance between the point on the parabola to the directrix

To find the equation of the parabola, equate these two expressions and solve for y0y0 .

Find the equation of the parabola in the example above.

Distance between the point (x0,y0)(x0,y0) and (a,b)(a,b) :

(x0−a)2+(y0−b)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√(x0−a)2+(y0−b)2

Distance between point (x0,y0)(x0,y0) and the line y=cy=c :

∣∣y0−c∣∣| y0−c |

(Here, the distance between the point and horizontal line is difference of their yy -coordinates.)

Equate the two expressions.

(x0−a)2+(y0−b)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=∣∣y0−c∣∣(x0−a)2+(y0−b)2=| y0−c |

Square both sides.

(x0−a)2+(y0−b)2=(y0−c)2(x0−a)2+(y0−b)2=(y0−c)2

Expand the expression in y0y0 on both sides and simplify.

(x0−a)2+b2−c2=2(b−c)y0(x0−a)2+b2−c2=2(b−c)y0

This equation in (x0,y0)(x0,y0) is true for all other values on the parabola and hence we can rewrite with (x,y)(x,y) .

Therefore, the equation of the parabola with focus (a,b)(a,b) and directrix y=cy=c is

(x−a)2+b2−c2=2(b−c)y

3 0

3 years ago

One of the vertices of an equilateral triangle is on the vertex of a square and two other vertices are on the not adjacent sides

Elina [12.6K]

<h2>Answer:</h2>

The side of the triangle is either 38.63ft or 10.35ft

<h2>Step-by-step explanation:</h2>

This problem can be translated as an image as shown in the Figure below. We know that:

The side of the square is 10 ft.
One of the vertices of an equilateral triangle is on the vertex of a square.
Two other vertices are on the not adjacent sides of the same square.

Let's call:

Since the given triangle is equilateral, each side measures the same length. So:

x: The side of the equilateral triangle (Triangle 1)

y: A side of another triangle called Triangle 2.

That length is the hypotenuse of other triangle called Triangle 2. Therefore, by Pythagorean theorem:

$\mathbf{(1)} \ x^2=100+y^2$

We have another triangle, called Triangle 3, and given that the side of the square is 10ft, then it is true that:

$y+(10-y)=10$

Therefore, for Triangle 3, we have that by Pythagorean theorem:

$(10-y)^2+(10-y)^2=x^2 \\ \\ 2(10-y)^2=x^2 \\ \\ \\ \mathbf{(2)} \ x^2=2(10-y)^2$

Matching equations (1) and (2):

$2(10-y)^2=100+y^2 \\ \\ 2(100-20y+y^2)=100+y^2 \\ \\ 200-40y+2y^2=100+y^2 \\ \\ (2y^2-y^2)-40y+(200-100)=0 \\ \\ y^2-40y+100=0$

Using quadratic formula:

$y_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ y_{1,2}=\frac{-(-40) \pm \sqrt{(-40)^2-4(1)(100)}}{2(1)} \\ \\ \\ y_{1}=37.32 \\ \\ y_{2}=2.68$

Finding x from (1):

$x^2=100+y^2 \\ \\ x_{1}=\sqrt{100+37.32^2} \\ \\ x_{1}=38.63ft \\ \\ \\ x_{2}=\sqrt{100+2.68^2} \\ \\ x_{2}=10.35ft$

Finally, the side of the triangle is either 38.63ft or 10.35ft

5 0

3 years ago

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