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3 years ago

Please help and no guessing please I need this.

Mathematics

1 answer:

vodka [1.7K]3 years ago

7 0

Answer:

1. n + 8 = 10

2. $\frac{n}{12}$ = 10

3. 8n - 6 = 34

4. 3(n + 12) = 45

5. No

6. Yes

Step-by-step explanation:

(only for questions 5 and 6)

5. If you substitute in y = 1, you get 1 * 49 = 0 or 49 = 0 which is not true.

6. If you substitute in b = 8, you get:

$5(8-3)= 25\\5(5) = 25\\25 = 25$

this is true.

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Instructions: Find the equation of the line through point (2, -1) and

Kazeer [188]

Answer:

$y = -\frac{2}{5}x -\frac{1}{5}$

Step-by-step explanation:

Given

Passes through: $(2,-1)$

Parallel to: $2x + 5y = 15$

Required

First, calculate the slope of the parallel line

$2x + 5y = 15$

Make y the subject

$5y = 15 - 2x$

Divide through by 5

$y = 3 - \frac{2}{5}x$

$y =- \frac{2}{5}x+3$

An equation in slope intercept has the form:

$y = mx + b$

Where:

m = slope

$y = mx + b$

So:

$m = -\frac{2}{5}$

The required is parallel to $2x + 5y = 15$ .

This mean, the same slope

The equation is the calculated using:

$y - y_1 = m(x - x_2)$

This gives:

$y - (-1) = -\frac{2}{5}(x - 2)$

$y +1 = -\frac{2}{5}x +\frac{2}{5}*2$

$y +1 = -\frac{2}{5}x +\frac{4}{5}$

$y = -\frac{2}{5}x +\frac{4}{5}-1$

Take LCM

$y = -\frac{2}{5}x +\frac{4-5}{5}$

$y = -\frac{2}{5}x -\frac{1}{5}$

i.e.

$y = -2/3x -1/5$

3 0

3 years ago

Can have both positive and negative oxidation numbers metal or nonmetal

LenaWriter [7]

Yes, they can have both positive and negative oxidation numbers.

4 0

3 years ago

Read the following statement:

Serga [27]

Well, since it says it's congruent, it's congruent. EF overbar congruent to GH overbar.

3 0

3 years ago

Read 2 more answers

The age of a father is 3 times the age of his son. 5 years ago, the father was 4

LUCKY_DIMON [66]

Answer:

Step-by-step explanation:

x = father's age

y = son's age

Nowadays:

3x = y

5 years ago:

4(x - 5) = y - 5

4x - 20 = y - 5

4x = y + 15

I hope I've helped you.

3 0

4 years ago

Inveres laplace transform (3s-14)/s^2-4s+8

Olenka [21]

Complete the square in the denominator.

$s^2 - 4s + 8 = (s^2 - 4s + 4) + 4 = (s-2)^2 + 4$

Rewrite the given transform as

$\dfrac{3s-14}{s^2-4s+8} = \dfrac{3(s-2) - 8}{(s-2)^2+4} = 3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}$

Now take the inverse transform:

$L^{-1}_t\left\{3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3L^{-1}_t\left\{\dfrac{s-2}{(s-2)^2+2^2}\right\} - 4L^{-1}_t\left\{\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3e^{2t} L^{-1}_t\left\{\dfrac s{s^2+2^2}\right\} - 4e^{2t} L^{-1}_t\left\{\dfrac{2}{s^2+2^2}\right\} \\\\ \boxed{3e^{2t} \cos(2t) - 4e^{2t} \sin(2t)}$

5 0

3 years ago