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3 years ago

The equation of a line is x = -4. Will this produce a horizontal, vertical, or diagnol line?

Mathematics

1 answer:

Nuetrik [128]3 years ago

3 0

Vertical because the line passed through -4 on the x axis

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Find the difference between -14w - 3 and 5w.

kirza4 [7]

(-14w) - (-3) - (-5w) equals
-9w + 3

or if what you're saying is
-14w - 3 - 5w it equals
-19w - 3

or if what youre saying is
(-14w - 3) - 5w it equals
-19w - 3

8 0

3 years ago

Read 2 more answers

How do you calculate viewing area

Sauron [17]

Answer:

The way to calculate viewing area would be base multiplied by height, but if you would be more specific I would provide a more in detail answer.

8 0

3 years ago

Mrs.Stone is three times older than her younger daughter, Joyce, but only two times older than her older daughter, Alex. How old

olchik [2.2K]

Answer:

54 years

Step-by-step explanation:

Joyce age = x

Alex age = y

3x = 2y - - - (1)

y - 9 = 2(x - 9)

y - 9 = 2x - 18

-9 + 18 = 2x - y

9 = 2x - y

y = 2x - 9 - - (2)

3x = 2y

3x = 2(2x - 9)

3x = 4x - 18

18 = 4x - 3x

18 = x

Hence,

Mrs stone is 3 * 18 = 54 years

5 0

2 years ago

A submarine at a depth of -25 2/3feet descended 5 1/3. How many feet must the submarine rise to reach the surface?

zheka24 [161]

Answer:

it descended 31 Feet b b b b b b b b b bb b b b b b b b b b b

7 0

3 years ago

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