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nalin [4]
3 years ago
7

John and Tammy cannot agree about the following statement: "If x = 9 and 4x - 7 = y, then 4(9) - 7 = y." John says this is the s

ubstitution property of equality and Tammy says it's the distributive property. Who is correct? Explain your reasoning.
Mathematics
1 answer:
oksian1 [2.3K]3 years ago
7 0
John is correct. The substitution property is being used. The x is being substituted for the value 9 (aka x is replaced with 9). Keep in mind that x is simply a placeholder for a number, in this case 9. 

Think of "substitution" as in a sports game where one player replaces another. Or you can think of a substitute teacher where someone else temporarily replaces your teacher for the day. Those are two ways to help remember the term.

The distributive property is not being used here since it is a*(b+c) = a*b+a*c. For example, 2*(3+5) = 2*3+2*5
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Rainbow [258]

Answer:

No units

Step-by-step explanation:

The cost function is given by;

C(x)=x^2-20x+35

To find the units manufactured at a cost of $5,560

We equate the C(x) to 5560 and solve for x.

x^2-20x+35 = 5560

x^2-20x+35 - 5560 = 0

x^2-20x - 5525= 0

The discriminant of this equation is negative.

That is,

d =  {( - 20)}^{2}  - 4 \times 1 \times  - 5560 \:  < 0

Hence the equation has no real roots.

This means no units can be produced at $5,560

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3 years ago
What is 0.72 as a percentage?
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2 years ago
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Rose has $30 to spend on school supplies. She spent a total of $12.52, including tax, on a new pack of binders. She needs to sav
zysi [14]

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3 years ago
Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without e
svet-max [94.6K]

ANSWER

y = 2x -4

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

x = 5 +  ln(t)

y =  {t}^{2}  + 5

From the first equation we make t the subject to get;

x - 5 =  ln(t)

t =  {e}^{x - 5}

We put it into the second equation.

y =  { ({e}^{x - 5}) }^{2}  + 5

y =  { ({e}^{2(x - 5)}) }  + 5

We differentiate to get;

\frac{dy}{dx}  = 2 {e}^{2(x - 5)}

At x=5,

\frac{dy}{dx}  = 2 {e}^{2(5 - 5)}

\frac{dy}{dx}  = 2 {e}^{0}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5)

y = 2x - 10 + 6

y = 2x -4

Without eliminating the parameter,

\frac{dy}{dx}  =  \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }

\frac{dy}{dx}  =  \frac{ 2t}{  \frac{1}{t} }

\frac{dy}{dx}  =  2 {t}^{2}

At x=5,

5 = 5 +  ln(t)

ln(t)  = 0

t =  {e}^{0}  = 1

This implies that,

\frac{dy}{dx}  =  2 {(1)}^{2}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5) =

y = 2x -4

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3 years ago
Val has 2 hours to finish a slide presentation. It takes her 1 ··6 hour to create each slide. Will Val have enough time to creat
Papessa [141]

Answer:

yes

Step-by-step explanation:

10 × 1/6 < 2

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3 years ago
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