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AlekseyPX
3 years ago
13

Cari owns a horse farm and a horse trailer than can transport up to 8000 pounds of livestock and tack. She travels with 5 horses

whose combined weight is 6240 pounds. let t represent the average weight of tack per horse. which of the following inequalities could be used to determine the weight of tack Cari can allow for each horse?
A. 6240+t≤8000
B. 8000≥6240t
C. 6240+5t≤8000
D. 8000+6240≤5t
E. 8000-6240≤5t
Mathematics
2 answers:
horsena [70]3 years ago
8 0

Answer: your answer should be D

Step-by-step explanation:

because the trailer is already 8000 pounds. so you would add that weight with the weight of the horses. the 5t would represent the number of horses and how many pounds they are since you don't know that part of the question.

Hdbdbfbfbfbf2 years ago
0 0

I think idk it’s b isjdnd

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X^+17x+72=12 factoring quadratic equation
Tom [10]

Answer:

The first term is, x2 its coefficient is 1 .

The middle term is, -17x its coefficient is -17 .

The last term, "the constant", is +60

Step-1 : Multiply the coefficient of the first term by the constant 1 • 60 = 60

Step-2 : Find two factors of 60 whose sum equals the coefficient of the middle term, which is -17 .

-60 + -1 = -61

-30 + -2 = -32

-20 + -3 = -23

-15 + -4 = -19

-12 + -5 = -17 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -12 and -5

x2 - 12x - 5x - 60

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (x-12)

Add up the last 2 terms, pulling out common factors :

5 • (x-12)

Step-5 : Add up the four terms of step 4 :

(x-5) • (x-12)

Which is the desired factorization

Equation at the end of step

1

:

(x - 5) • (x - 12) = 0

STEP

2

:

Theory - Roots of a product

2.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation:

2.2 Solve : x-5 = 0

Add 5 to both sides of the equation :

x = 5

Solving a Single Variable Equation:

2.3 Solve : x-12 = 0

Add 12 to both sides of the equation :

x = 12

Supplement : Solving Quadratic Equation Directly

Solving x2-17x+60 = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex:

3.1 Find the Vertex of y = x2-17x+60

For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 8.5000

Plugging into the parabola formula 8.5000 for x we can calculate the y -coordinate :

y = 1.0 * 8.50 * 8.50 - 17.0 * 8.50 + 60.0

or y = -12.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2-17x+60

Axis of Symmetry (dashed) {x}={ 8.50}

Vertex at {x,y} = { 8.50,-12.25}

x -Intercepts (Roots) :

Root 1 at {x,y} = { 5.00, 0.00}

Root 2 at {x,y} = {12.00, 0.00}

Solve Quadratic Equation by Completing The Square

3.2 Solving x2-17x+60 = 0 by Completing The Square .

Subtract 60 from both side of the equation :

x2-17x = -60

Now the clever bit: Take the coefficient of x , which is 17 , divide by two, giving 17/2 , and finally square it giving 289/4

Add 289/4 to both sides of the equation :

On the right hand side we have :

-60 + 289/4 or, (-60/1)+(289/4)

The common denominator of the two fractions is 4 Adding (-240/4)+(289/4) gives 49/4

So adding to both sides we finally get :

x2-17x+(289/4) = 49/4

Adding 289/4 has completed the left hand side into a perfect square :

x2-17x+(289/4) =

(x-(17/2)) • (x-(17/2)) =

(x-(17/2))2

Things which are equal to the same thing are also equal to one another. Since

x2-17x+(289/4) = 49/4 and

x2-17x+(289/4) = (x-(17/2))2

then, according to the law of transitivity,

(x-(17/2))2 = 49/4

We'll refer to this Equation as Eq. #3.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x-(17/2))2 is

(x-(17/2))2/2 =

(x-(17/2))1 =

x-(17/2)

Now, applying the Square Root Principle to Eq. #3.2.1 we get:

x-(17/2) = √ 49/4

Add 17/2 to both sides to obtain:

x = 17/2 + √ 49/4

Since a square root has two values, one positive and the other negative

x2 - 17x + 60 = 0

has two solutions:

x = 17/2 + √ 49/4

or

x = 17/2 - √ 49/4

Note that √ 49/4 can be written as

√ 49 / √ 4 which is 7 / 2

Solve Quadratic Equation using the Quadratic Formula

3.3 Solving x2-17x+60 = 0 by the Quadratic Formula .

According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

- B ± √ B2-4AC

x = ————————

2A

In our case, A = 1

B = -17

C = 60

Accordingly, B2 - 4AC =

289 - 240 =

49

Applying the quadratic formula :

17 ± √ 49

x = —————

2

Can √ 49 be simplified ?

Yes! The prime factorization of 49 is

7•7

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 49 = √ 7•7 =

± 7 • √ 1 =

± 7

So now we are looking at:

x = ( 17 ± 7) / 2

Two real solutions:

x =(17+√49)/2=(17+7)/2= 12.000

or:

x =(17-√49)/2=(17-7)/2= 5.000

Two solutions were found :

x = 12

x = 5

Step-by-step explanation:

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