Your local radio station is having their yearly music player and concert ticket giveaway. For one minute, every 5^\text{th}5 th 5, start superscript, t, h, end superscript caller will win a music player and every 7^\text{th}7 th 7, start superscript, t, h, end superscript caller will win concert tickets. You were just the first caller to win both a music player and concert tickets! What number caller were you?

Hello,

To solve we need to factor y^2-2y-35:

(y-7)(y+5)=0

Now solve for :

(When will (y-7)(y+5)equal zero?)

When y-7=0 or y+5=0

Solve for the 2 equations above and;

y= 7,-5

Hope this helps, have a great day!

Hey there!!

Given equation :

d(e-f) = g

Solve for d :

Dividing by ( e - f ) on both sides :

d = g / e - f which is [tex]d=\frac{g}{e-f}[/tex]

Hope my answer helps!

There was no change in Samuel's savings account balance. He made a deposit of $160 but after two weeks, he withdrew the same sum. He still had $20 in his account as a result.

This math query relates to fundamental financial operations. Samuel has $20 in his savings account at the start of the story. His total now stands at $180 after his subsequent $160 deposit. He withdraws $160, though, after two weeks. He now has $20 in his account after that. Finally, there was no change in Samuel's savings account balance. His account was $20 from beginning to end.

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probability has 11 letters

2/11 chances that you could get a B

5/9 for a T or vowel

3/11 for an O or B

you just look at the letters and the number of them

Answer:

B

Step-by-step explanation:

Therefore a geometric figure labeled CD would be a segment and the correct answer is B)

Answer:

The Coordinate of the vertices of Parallelogram RSTU are R(0,0), S(2,3),T(6,3) and U (4,0).

we have to find the vertices of a point which cuts the side of parallelogram ST  .It is given that line y = x passes through R.

Suppose that it cuts the side ST at M (p,q).

Equation of line passing through T(6,3)and S (2,3) is

[tex]\frac{y-3}{x-2}= \frac{3-3}{6-3}\\\frac{y-3}{x-2} = 0\\y- 3=0[/tex]

it passes through (p,q)∴

∴ q -3 =0     ...........(1)

Line y=x passes through (p,q).

p -q=0

p=q

Equation (1) becomes

p= q=3

So the line y =x cuts TU at M (3,3).

$6 dollars because 11-8 = 3, 3x2 =6

He starts with $6.00

It says he spends 1/2 of his allowance on the movies. 6/2=3

He has $3. Then it says He gained $8.00 for washing the "family" car.

$6.00+$8.00

=$11

Answer=$6.00

Every year there are 14 more ducks. The starting point would be 50 and the equation is:

d = 14*year+50

Hello,

Please, see the attached files.

Thanks.

Answer:

It is given that Graph the image of the given triangle after the transformation that has the rule (x, y)→(−x, −y) .

From the given figure it is noticed that the vertices of the triangle are (6,7), (7,4), (-2,5).

From the given rule (x, y)→(−x, −y)  the vertices of the image are (-6,-7), (-7,-4) and (2,-5).

Plot these points on the coordinate plane.

In the below graph, the image of the given triangle is represented by green lines.

24 = 6kx

4 = kx

4/k = x

Let's solve for x.

3kx+24=9kx

Step 1: Add -9kx to both sides.

3kx+24+−9kx=9kx+−9kx

−6kx+24=0

Step 2: Add -24 to both sides.

−6kx+24+−24=0+−24

−6kx=−24

Step 3: Divide both sides by -6x.

[tex]\frac{-6kx}{-6x}[/tex] = [tex]\frac{-24}{-6k}[/tex]

[tex]x=\frac{4}{k}[/tex]

Answer is: 4/k

The price of the stock of PEV Corporation varies as shown. we have to determine the net charge. Net change is just how much it changed overall, so we will look for all the ups minus all the downs.  

Here, "went up" goes with a + sign, and "went down" goes with a - sign.

So, Net charge = [tex]\frac{1}{4} - \frac{1}{3}-\frac{3}{4} + \frac{7}{10}[/tex]

= [tex]\frac{1}{4}-\frac{3}{4}- \frac{1}{3} + \frac{7}{10}[/tex]

= [tex]-\frac{2}{4}- \frac{1}{3} + \frac{7}{10}[/tex]

= [tex]-\frac{1}{2}- \frac{1}{3} + \frac{7}{10}[/tex]

LCM of 2,3 and 10 is 30

= [tex]\frac{-15-10+21}{30}[/tex]

= [tex]\frac{-4}{30}[/tex]

= [tex]\frac{-2}{15}[/tex]

So, the net charge is [tex]\frac{-2}{15}[/tex]

​(x²-y²)² + (2xy)² = (x²+y²)²

Find the missing x- and y-values and Pythagorean triples using the identity given

A Pythagorean triple consists of three positive integers a, b, and c, that satisfy the equation from the Pythagorean theorem, thus, a² + b² = c², such triple is commonly written (a,b,c).​

​We are given the equation : (x²-y²)² + (2xy)² = (x²+y²)² since this, we have :

​a = (x²-y²)

​b = (2xy)

​c = (x²+y²)

​Question 1)

X Value = 4​

​Y Value = 3

​Pythagorean triples: ?

We can replace the values of x and y, to determine a, b and c.​

​a = (x²-y²) = (4²-3²) = 16-9 = 7

​b = (2xy) = (2*4*3) = 24

c = (x²+y²) = (4²+3²) = 16+9 = 25​

​Answer 1 : Pythagorean triples : (7,24,25)

Question 2) ​

​X Value = 5

Y Value = ?​

Pythagorean Triples: (9,40,41)​

​Now we have a, b, and c, to determine Y

​b = (2xy) = 40

Y = 40/2x = 40/2*5 = 40/10 = 4​

Answer 2 : Y = 4​

​Question 3)

X Value = ?​

Y Value = 3​

Pythagorean Triples: (27,36,45)​

Now we have a, b, and c, to determine X​

b = (2xy) = 36​

​X = 36/2y = 36/2*3 = 36/6 = 6

​Answer 3 : X = 6

Question 4)​

​X Value = 7

Y Value = 5​

​Pythagorean Triples: ?

We can replace the values of x and y, to determine a, b and c.​

a = (x²-y²) = (7²-5²) = 49-25 = 24​

b = (2xy) = (2*7*5) = 70​

c = (x²+y²) = (7²+5²) = 49+25 = 74​

Answer 4 : Pythagorean triples : (24,70,74)​

Hope this helps!​​​​​

​[tex]\textit{\textbf{Spymore}}[/tex]​​​​​​​

By using the given identity and substituting known values of x and y, we can find missing x or y values and compute the Pythagorean triples. For instance, when x=4 and y=3, the Pythagorean triple is (7,24,25). Similarly, we can find missing triples or variables for other sets.

This question pertains to the topic of Pythagorean triples and their relationship with a given identity. A Pythagorean triple consists of three positive integers (a, b, c), such that a2 + b2 = c2. Inserting known values of x and y into the identity provided can help us find the Pythagorean triples.

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Answer:

1)  a=12 , b=-9 , c=4

2)  a=-1 , b=-10 , c=2

3)  a=5 , b=-1 , c=9

4)  a=1 , b=-8 , c=16

Step-by-step explanation:

(1) y = 12x² - 9x + 4 no real solutions.

(2) 10x + y = -x² + 2, two real solutions.

(3) 4y - 7 = 5x² - x + 2 + 3y, no real solution.

(4) y = (-x + 4)², one real solution.

The number of real solutions for each quadratic equation is determined using discriminant.

The discriminant, denoted as Δ, is used to determine the nature of the roots of a quadratic equation of the form ax² + bx + c = 0.

Δ = b² - 4ac

From the given equations;

y = 12x² - 9x + 4

a = 12, b = -9, c = 4

Δ = (-9)² - 4(12)(4)

Δ = 81 - 192

Δ = -111

Since the discriminant Δ is negative (-111), this quadratic equation has no real solutions.

10x + y = -x² + 2

y = -x² - 10x + 2

a = -1, b = -10, c = 2

Δ = (-10)² - 4(-1)(2)

Δ = 100 + 8

Δ = 108

Since the Δ is positive and greater than zero, the equation has two real solutions.

4y - 7 = 5x² - x + 2 + 3y

4y - 3y = 5x² - x + 2 + 7

y = 5x² - x + 9

a = 5, b = -1, c = 9

Δ = (-1)² - 4(5)(9)

Δ = 1 - 20(9)

Δ = 1 - 180 = -179

No real solution.

y = (-x + 4)²

when y = 0

0 = -x + 4 or

0 = -x + 4

x = 4

The equation has one real solution.

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Answer:

Given: [tex]m\angle 1 = m\angle 3[/tex] and [tex]m\angle 2 = m\angle 3[/tex]

To prove that:

[tex]l || m[/tex]

1. [tex]m\angle 1 = m\angle 3[/tex]          [Given]

[tex]m\angle 2 = m\angle 3[/tex]

Substitution property of equality says that:

If x = y, then x can be substituted in y, or y can be substituted in x.

2 [tex]m\angle 1 = m\angle 2[/tex]          [ By Substitution Property]

Alternate interior angles states that when two lines are crossed by transversal , a pair of angles on the inner sides of each of these two lines on the opposite sides of the transversal line.

3. [tex]\angle 1[/tex] and  [tex]\angle 2[/tex] are alternate interior angles  [By definition Alternate interior angle].

Alternating interior angles theorem states that if two parallel lines are intersected by third lines, then the angles in the inner sides of the parallel lines on the opposite sides of the transversal are equal.

4. [tex]l || m[/tex] ; then the lines are parallel     [By Alternate interior angles theorem]

Correct match is as follows:

1. [tex]m\angle 1 = m\angle 3[/tex]          [Given]

 [tex]m\angle 2 = m\angle 3[/tex]

2. [tex]m\angle 1 = m\angle 2[/tex]          [Substitution]

3. [tex]\angle 1[/tex] and  [tex]\angle 2[/tex] are alternate interior angle       [By definition of alternate interior angles ]

4. [tex]l || m[/tex] the lines are parallel  [If alternate interior angles are equal]

The question involves a geometrical proof where measurements of angles are given and a conclusion regarding parallel lines need to be derived. The steps of proof are matched with relevant geometric concepts. The matching involves principles like substitution, definition of alternate interior angles, and a theorem about alternate interior angles and parallel lines.

In the context of the given problem, the matching would be as follows:

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To determine the approximate number of workers hired during the seventh year based on the quadratic model, we need more information about the model and the values of a, b, and c.

To approximate the number of workers hired during the seventh year based on the quadratic model, we need more information about the model. Quadratic models are typically of the form y = ax^2 + bx + c, where x represents the time and y represents the number of workers hired. Without knowing the values of a, b, and c, we cannot determine the exact number of workers hired in the seventh year. However, if we have the values of a, b, and c, we can substitute x = 7 into the equation to find the approximate number of workers hired during the seventh year.

Answer:

$96

Step-by-step explanation:

Multiply The Number Of Tickets By Amount Of People Going

16 * 6

[tex]x^4-625=(x^2-25)(x^2+25)=(x-5)(x+5)(x-5i)(x+5i)\\\\x^4+10x^2+25=(x^2+5)^2=((x-\sqrt5i)(x+\sqrt5i))^2=(x-\sqrt5i)^2(x+\sqrt5i)^2[/tex]

Answer:

1) [tex]x^4-625=(x+5i)(x-5i)(x+5)(x-5)[/tex]

2) [tex]x^4+10x^2+25=(x+\sqrt{5}i)^2(x-\sqrt5 i)^2[/tex]

Step-by-step explanation:

1) Given : Expression [tex]x^4-625[/tex]

To find : Factor the expression completely over the complex numbers ?

Solution :

We can re-write the expression as,

[tex]x^4-625=(x^2)^2-(25)^2[/tex]

Applying identity, [tex]a^2-b^2 = (a+b)(a-b)[/tex]

[tex]x^4-625=(x^2+25)(x^2-25)[/tex]

[tex]x^4-625=(x^2+25)(x^2-5^2)[/tex]

Again apply same identity,

[tex]x^4-625=(x^2+25)(x+5)(x-5) [/tex]

The factor of [tex]x^2+25=(x+5i)(x-5i)[/tex]

Factor form is [tex]x^4-625=(x+5i)(x-5i)(x+5)(x-5)[/tex]

2) Given : Expression [tex]x^4+10x^2+25[/tex]

To find : Factor the expression completely over the complex numbers ?

Solution :

Expression [tex]x^4+10x^2+25[/tex]

Let [tex]x^2=y[/tex]

[tex]y^2+10y+25[/tex]

To factor we equate it to zero.

[tex]y^2+10y+25=0[/tex]

Apply middle term split,

[tex]y^2+5y+5y+25=0[/tex]

[tex]y(y+5)+5(y+5)=0[/tex]

[tex](y+5)(y+5)=0[/tex]

Substitute back,

[tex](x^2+5)(x^2+5)=0[/tex]

[tex](x^2+5)^2=0[/tex]

[tex]x^2+5=0[/tex]

[tex]x^2=-5[/tex]

Taking root both side,

[tex]x=\sqrt{-5}[/tex]

[tex]x=\pm \sqrt{5}i[/tex]

So, The factors are [tex](x+\sqrt{5}i)(x-\sqrt5 i)[/tex]

Factor form is [tex]x^4+10x^2+25=(x+\sqrt{5}i)^2(x-\sqrt5 i)^2[/tex]

The slope formula:

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

We have the points (4, 9) and (6, 1).

Substitute

[tex]m=\dfrac{1-9}{6-4}=\dfrac{-8}{2}=-4[/tex]

x = 6 or x = - 1

rearrange the equation into standard form and factorise

subtract ( - 2x - 4 ) from both sides

x² - 5x - 6 = 0 ← in standard form

the factors of - 6 which sum to - 5 are - 6 and + 1

(x - 6)(x + 1) = 0

equate each factor to zero and solve for x

x - 6 = 0 ⇒ x = 6

x + 1 = 0 ⇒ x = - 1

Answer: Answer would be 23

Step-by-step explanation:

Suppose that Carol has x 10p coins in her purse. Since she has three times as many 20p coins as 10p coins, then she has 3x 20p coins in her purse.

x coins of 10p value $0.10x and 3x coins of 20p value $0.60x.

In total this amount of money is $(0.10x+0.60x)=$0.70x that is exactly $4.20. Thus, you get the equation

0.70x=4.20.

Solve it:

[tex]x=\dfrac{4.20}{0.70}=6.[/tex]

Carol has 6 coins of 10p and 3·6=18 coins of 20p.

Answer:

I'm not sure how euros work but I'm almost certain there are 18 20p coins

Step-by-step explanation:

y = x^2 to  Green

y = –2(x – 2)^2 + 2  Blue

The correct answer is D.

Answer:

The correct option is D.

Step-by-step explanation:

The parent function is

[tex]y=x^2[/tex]

The transformed function is

[tex]y=-2(x-2)^2+2[/tex]              .... (1)

The translation is defined as

[tex]f(x)=k(x+a)^2+b[/tex]                .... (2)

Where, k is stretch factor, a is horizontal shift and b is vertical shift.

If a>0, then the graph shifts a units left and if a<0, then the graph shifts a units right.

If b>0, then the graph shifts b units up and if b<0, then the graph shifts b units down.

From (1) and (2) it is clear that

[tex]a=-2,b=2,k=-2[/tex]

It means,

a=-2<0 the graph shifts 2 units right.

b=2>0 the graph shifts 2 units up.

k=-2, the graph reflect across the x-axis and stretch by the factor 2.

The graph reflect across the x-axis, translate 2 units to the right, translate up 2 units, stretch by the factor 2. Therefore the correct option is D.

Answer:

1668

Step-by-step explanation:

8(8)= 64

64+4=68

1,600+68= 1,668

729 = s^3

9^3 = s^3

s = 9 in.

Answer:

B because eaxh section either repeats itself or is close to repeating its previous part

Step-by-step explanation:

$90/10 = 9

$160/20 = 8

$80/10 = 10

$120/20 = 6

The package that has the lowest ratio cost to number of lessons is $120 for 20 lessons.

Hope this Helps!!

The package with the lowest ratio of cost to number of lessons is the $120 for 20 lessons option, calculating to a cost of $6 per lesson, which is the most cost-effective option.

To find this, we calculate the cost per lesson for each package by dividing the total cost by the number of lessons.

Therefore, the package with the lowest ratio of cost to number of lessons is $120 for 20 lessons, with a cost of $6 per lesson.