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3 years ago

Write the vertex form equivalent of 3x^2-30x+73=y

Mathematics

1 answer:

Naddik [55]3 years ago

5 0

Answer:

(5,-2) or y=3(x-5)^2 -2

Step-by-step explanation:

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HELP ME ASAP!!!!!!! IT IS QUICK PROMISE JST ANSWER TWO SIMPLE QUESTIONS

Viefleur [7K]

Answer:

3.6 for tomatoes and 8.4 for lettuce

Step-by-step explanation:

70% of 12=8.4

30% of 12=3.6

check: 3.6+8.4=12

4 0

2 years ago

A quadratic equation is shown below: 3x^2 − 15x + 20 = 0 Part A: Describe the solution(s) to the equation by just determining th

Anuta_ua [19.1K]

These are two questions and two answers:

Question 1:

A quadratic equation is shown below: 3x^2 − 15x + 20 = 0 Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work.

Answer: The negative value of the radicand means that the equation does not have real solutions.

Explanation:

1) With radicand the statement means the disciminant of the quadratic function.

2) The discriminant is: b² - 4ac, where a, b, and c are the coefficients of the quadratic equation: ax² + bx + c

3) Then, for 3x² - 15x + 20, a = 3, b = - 15, and c = 20

and the discriminant (radicand) is: (-15)² - 4(3)(20) = 225 - 240 = - 15.

4) The negative value of the radicand means that the equation does not have real solutions.

Question 2:

Part B: Solve 3x^2 + 5x − 8 = 0 by using an appropriate method. Show the steps of your work, and explain why you chose the method used.

Answer: two solutions x = 1 and x = - 8/3x

Explanation:

1) I choose factoring (you may use the quadratic formula if you prefer)

2) Factoring

Given: 3x² + 5x − 8 = 0

Make 5x = 8x - 3x: 3x² + 8x - 3x - 8 = 0

Group: (3x² - 3x) + (8x - 8) = 0

Common factors for each group: 3x(x -1) + 8(x - 1) = 0

Coomon factor x - 1: (x - 1) (3x + 8) = 0

The two solutions are for each factor equal to zero:

x - 1 = 0 ⇒ x = 1
3x + 8 = 0 ⇒ x = -8/3

Those are the two solutions. x = 1 and x = - 8/3

4 0

2 years ago

Write an equation for the statement: A kilogram is equal to approximately 2.2 pounds. A cinder block weighs x pounds. A brick ha

navik [9.2K]

the total mass in kilograms y of the cinder block and brick is $y=0.45x + 0.6$ .

Step-by-step explanation:

Here we have , A kilogram is equal to approximately 2.2 pounds. A cinder block weighs x pounds. A brick has a mass of 0.6 kilogram. We need to find What is the total mass in kilograms y of the cinder block and brick . Let's find out :

According to given data in question ,

⇒ $1kg = 2.2pounds$

⇒ $1(pounds) = \frac{1}{2.2}kg$

A cinder block weighs x pounds and a brick has a mass of 0.6 kilogram . Let total mass in kg of the cinder block and brick is y , i.e.

⇒ $y (kg)= x (pounds) + 0.6 kg$

⇒ $y (kg)= x (\frac{1}{2.2}kg) + 0.6 kg$

⇒ $y=0.45x + 0.6$

Therefore , the total mass in kilograms y of the cinder block and brick is $y=0.45x + 0.6$ .

7 0

2 years ago

Write a polynomial equation of least degree with roots 2, -1 and 4.

Rasek [7]

A polynomial with roots a and b is (x - a)(x - b).

(x - 2)(x - (-1))(x - 4) = 0

(x - 2)(x + 1)(x - 4) = 0

has roots 2, -1, and 4.

3 0

2 years ago

For the cost function c equals 0.1 q squared plus 2.1 q plus 8, how fast does c change with respect to q when q equals 11? Det

Soloha48 [4]

Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is 9.95%

Step-by-step explanation:

Cost function is given as, $c=0.1\:q^{2}+2.1\:q+8$

Given that c changes with respect to q that is, $\dfrac{dc}{dq}$ . So differentiating given function,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)$

Applying sum rule of derivative,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)$

Applying power rule and constant rule of derivative,

$\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0$

$\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1$

$\dfrac{dc}{dt}=0.2\left(q\right)+2.1$

Substituting the value of $q=11$ ,

$\dfrac{dc}{dt}=0.2\left(11\right)+21.$

$\dfrac{dc}{dt}=2.2+2.1$

$\dfrac{dc}{dt}=4.3$

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,

$Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100$

Rewriting in terms of cost C,

$Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100$

Calculating value of $C\left(q \right)$

$C\left(q\right)=0.1\:q^{2}+2.1\:q+8$

Substituting the value of $q=11$ ,

$C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8$

$C\left(q\right)=0.1\left(121\right)+23.1+8$

$C\left(q\right)=12.1+23.1+8$

$C\left(q\right)=43.2$

Now using the formula for percentage,

$Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100$

$Percentage\:rate\:of\:change=0.0995\times 100$

$Percentage\:rate\:of\:change=9.95%$

Percentage rate of change of c with respect to q is 9.95%

7 0

2 years ago