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2 years ago

What is the final cost for an item that cost $3000 and has a tax rate of 5%? 50

Mathematics

1 answer:

lubasha [3.4K]2 years ago

7 0

Answer:

Sales tax: $150.00

■ Cost/Price before ST: $3,000.00

■ Total Cost/Price including ST: $3,150.00

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Parallel / Perpendicular Practice

deff fn [24]

The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

Neither
║
Neither
⊥
║
Neither
Neither
Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; $m_1 = -\dfrac{1}{m_2}$

1. The given equations are in the slope and intercept form

$\ y = 3 \cdot x + 1$

The slope, m₁ = 3

$y = \dfrac{1}{3} \cdot x + 1$

The slope, m₂ = $\dfrac{1}{3}$

Therefore, the equations are neither parallel or perpendicular

Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

$y = 5 \cdot x -\dfrac{7}{2}$

Therefore, the two equations are parallel

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

$y = 2 -\dfrac{1}{2} \cdot x$

Slope, m₁ = $-\dfrac{1}{2}$

$y = \dfrac{1}{2} \cdot x - 2$

Slope, m₂ = $\dfrac{1}{2}$

The slopes

Therefore, m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

The lines are Neither parallel nor perpendicular

Neither

4. The given equations are;

2·y - x = 2

$y = \dfrac{1}{2} \cdot x +1$

m₁ = $\dfrac{1}{2}$

y = -2·x + 4

m₂ = -2

Therefore;

$m_1 \neq -\dfrac{1}{m_2}$

Therefore, the lines are perpendicular

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, $y = \dfrac{3}{4} \cdot x + 3$

Second equation, $y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}$

Therefore, m₁ = m₂, the lines are parallel

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = $\dfrac{13}{5}$

Therefore, the two equations are neither parallel nor perpendicular

Neither

7. The equations are;

2·x + 6·y = -3

Which gives $y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}$

12·y = 4·x + 20

Which gives

$y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}$

m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

Neither

8. 2·x - 5·y = -3

Which gives; $y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}$

5·x + 27 = 6

$x = -\dfrac{21}{5}$

Therefore, the slopes are not equal, or perpendicular, the correct option is Neither

Learn more here:

brainly.com/question/16732089

6 0

3 years ago

I took too many pills counting up the bills Molly in my cup i can't tell you how it feel

AveGali [126]

Answer:

errday i b countin up countin up da blues count away sometimes we always lose

Step-by-step explanation:

juice wrld

5 0

3 years ago

Read 2 more answers

Show how you can use repeated subtraction to find 84÷6

Ganezh [65]

84-14-14-14-14-14-14

7 0

3 years ago

Read 2 more answers

Please help !!

shutvik [7]

Given:

The limit problem is:

$\lim_{x\to 3}(x^2+8x-2)$

To find:

The limit of the function by using direct substitution.

Solution:

We have,

$\lim_{x\to 3}(x^2+8x-2)$

Applying limit, we get

$\lim_{x\to 3}(x^2+8x-2)=(3)^2+8(3)-2$

$\lim_{x\to 3}(x^2+8x-2)=9+24-2$

$\lim_{x\to 3}(x^2+8x-2)=33-2$

$\lim_{x\to 3}(x^2+8x-2)=31$

Therefore, the correct option is D.

4 0

3 years ago

4x + 2(x + 8) = -19+ x

Likurg_2 [28]

4x + 2(x + 8) = -19 + x

4x + 2x + 16 = -19 + x

6x + 16 = -19 + x

6x + 35 = x

-5x = 35

x = -7

6 0

3 years ago

Read 2 more answers