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3 years ago

Determine the discount between the point(-4,2) and the line 4y=3×+6

Mathematics

1 answer:

frozen [14]3 years ago

6 0

Answer:

d = 14/5

Step-by-step explanation:

The point (-4,2) means that;

At x = -4, y = 2

Now general form of a linear equation is;

Ax + By + C = 0

We are given;

4y = 3x + 6

Rearranging to the form of a linear equation gives;

3x - 4y + 6 = 0

Thus, A = 3, B = -4 and C = 6

Thus, at point (-4,2), distance between them is;

d = (3(-4) - 4(2) + 6)/√(3² + (-4)²)

d = -14/5

We will take the absolute value.

Thus; d = 14/5

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The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

3 years ago

He cost, in millions of dollars, of building a three-story high school in new york state was estimated to be c(x) = 2.1 + 0.12x

ratelena [41]

Given cost function is

c(x) = $2.1 + 0.12x - 0.0001x^2$ (20 ≤ x ≤ 400)

where x is the number of thousands of square feet

total revenue will be $0.2 million dollars per thousand square feet

Revenue is 0.2 millions per thousand square feet. we know x is the number of thousand of square feet

So R(x) = 0.2x

We know Profit = Revenue - Cost

P(x) = R(x) - C(x)

$P(x) = 0.2x - (2.1 + 0.12x - 0.0001x^2)$

$P(x) = 0.2x - 2.1 - 0.12x + 0.0001x^2)$

combine like terms

$P(x) =-2.1 + 0.08x + 0.0001x^2$

Profit function is

$P(x) =0.0001x^2 + 0.08x - 2.1$

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3 years ago

Determine the discount between the point(-4,2) and the line 4y=3×+6​

Determine the discount between the point(-4,2) and the line 4y=3×+6