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3 years ago

Find the slope of the line in simplest form.

Mathematics

1 answer:

N76 [4]3 years ago

7 0

Answer:

$-\frac{3}{2}$

Step-by-step explanation:

To find the slope of the given line, slope is the rise divided by the run of a line.

Slope = $\frac{y_{2} - y_{1} }{x_{2} - x_{1} }$

Picking points (4, 18) and (8, 12)

Now;

x₁ = 4

y₁ = 18

x₂ = 8

y₂ = 12

Slope = $\frac{12 - 18}{8-4}$ = $-\frac{6}{4}$ = $-\frac{3}{2}$

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The purchase price of a car is $22,500.00 and the 5 year loan has a 10% down payment and an annual interest rate of 6.75%. What

kozerog [31]

Answer:

$398.59

Step-by-step explanation:

7 0

3 years ago

A water tank is leaking such that for each of the next 5 hours, the amount of water in the tank will be 2 gallons less than at t

andre [41]

Answer:

C. Linear decrease

Step-by-step explanation:

I can't really show a graph but I will try to explain

The water is leaking at a constant rate, meaning its not regaining it water (its only leaking). So if u were to graph this it would be a linear decrease

( That was a Terrible explanation but the answer is right lol)

7 0

3 years ago

The manufacturer offers a $125 rebate on photocopiers purchases before january 1. what is the nest cost of a $1,475 photocopier

Yuri [45]

Answer:

The net cost of the photocopier before january 1 is <u>$1350.</u>

Step-by-step explanation:

Given:

The manufacturer offers a $125 rebate on photocopiers purchases before january 1.

Now, to find the net cost of $1,475 photocopier before january 1.

Amount of rebate before January 1 = $125.

Cost of photocopier = $1,475.

So, to get the net cost of $1475 photocopier before january 1 we subtract the rebate from it:

$\$1475-\$125$

$=\$1350.$

Therefore, the net cost of the photocopier before january 1 is $1350.

6 0

3 years ago

Is 6,8,9 a pythagorem theorem? pls help me due rn

Naddika [18.5K]

Answer:

No, it isn't.

Step-by-step explanation:

Pythagorean theorem is x^2 + y^2 = z^2.

6^2 + 8^2 = 9^2

36 + 64 = 81

100 = 81

So, it isn't.

4 0

2 years ago

Read 2 more answers

Rationalise the denominator of:<br>1/(√3 + √5 - √2)

Paul [167]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} }$

can be re-arranged as

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} - \sqrt{2} + \sqrt{5} }$

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} } \times \dfrac{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this, we get

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ {( \sqrt{3} - \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{3 + 2 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{5 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{ - ( - \sqrt{3} + \sqrt{2} + \sqrt{5}) }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}}$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}} \times \dfrac{ \sqrt{6} }{ \sqrt{6} }$

$\rm \: = \: \dfrac{- \sqrt{18} + \sqrt{12} + \sqrt{30}}{2 \times 6}$

$\rm \: = \: \dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}$

$\rm \: = \: \dfrac{- 3\sqrt{2} + 2 \sqrt{3} + \sqrt{30}}{12}$

Hence,

$\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} } =\dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}}}$

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<h3><u>More Identities to </u><u>know:</u></h3>

$\purple{\boxed{\tt{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}}$

$\purple{\boxed{\tt{ {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}}$

6 0

3 years ago