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3 years ago

2) Write an equation that passes through the point (-2,5) and is parallel to y ==x+1.

Mathematics

1 answer:

Marta_Voda [28]3 years ago

4 0

Answer:

y = x + 7

Step-by-step explanation:

5 = -2 + b

7 = b

y = x + 7

* Parallel lines have SIMILAR RATE OF CHANGES [SLOPES], so 1 remains the way it is.

I am joyous to assist you anytime.

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The value of a car depreciates by 40% each year. At the end of 2007 the value of the car was £3600 Work out the value of the car

natima [27]

Answer: $6,000

Step-by-step explanation: If set up correctly, the value of the car at the end of 2006 times 0.6 (because the car is losing 0.4 of its value, so it keeps 0.6 of its value) should get you the value of the car at the end of 2007.

The equation should look like this:

(value of the car at the end of 2006) x 0.6 = (value of the car at the end of 2007)

Then you substitute: (value at the end of 2006) x 0.6 = 3600

Then divide: (value at the end of 2006) = 0.6 ÷ 3600

Then substitute again: (value at the end of 2006) = 6,000

Hope this helped you! Feel free to ask me any questions!

3 0

3 years ago

Whats a quadrilateral with two sides going in the same direction

Mariana [72]

Answer:

Trapezoid.

Step-by-step explanation:

That is a trapezoid ( in the UK a trapezium).

There is a pair of parallel sides.

6 0

2 years ago

1. Evaluate the expression if y = 5 and z = 0

natta225 [31]

Answer:

1. 0

2. -1

3. $120

4. $653.25

5. 4ab + 7a - 2b

Step-by-step explanation:

1. y = 5, z = 0

2z/y = ?

You'll need to substitute the values of y and z in the expression

2z/y

Becomes

2 * 0 / 5

= 0

So, 2z/y equals 0

......................................................................................................................

2. x = 0, y = -1 z = 7

xz - y²

Substituting the values of x, y and z in the above algebraic expression

xz - y²

Becomes

0*7 - (-1²)

0 - 1

-1

So, xz - y² equals -1

......................................................................................................................

3. The question falls under SIMPLE INTEREST

The formula is

I = (PRT)/100

Where I represents Interest

P = Principal (Amount Invested)

R = Rate of Interest

T = Time (in years)

From the question,

The interest is unknown ----------------- (To be calculated)

P equals $600

R = 2%

T = 10 years

Substitute these values in the formula above

I = (PRT)/100

Becomes

I = (600 * 2 * 10) / 100

I = 12000/100

I = 120

Result: The interest is $120

......................................................................................................................

The formula is

I = (PRT)/100

Where I represents Interest

P = Principal (Amount Invested)

R = Rate of Interest

T = Time (in years)

From the question,

The interest is unknown ----------------- (To be calculated)

P equals $13000

R = 6.7%

T = 9 months

Remember that time has to be in years

So, 9 months must be converted to year(s)

If 12 months makes a years

Then 9 months would be 9/12 years

So, T = 9/12 years

T = 0.75 years

Substitute these values in the formula above

I = (PRT)/100

I = (13000 * 6.7 * 0.75) / 100

I = 65325/100

I = 653.25

Result: The interest is $653.25

5. Simplify 4ab - 3bc + 7a - 2b + 3bc

4ab - 3bc + 7a - 2b + 3bc---------------------- Collect like terms

4ab + 7a - 2b + 3bc - 3bc

4ab + 7a - 2b

6 0

3 years ago

What is the X-coordinate of the Point (6,8)?

klio [65]

6 is the x coordinte

5 0

2 years ago

Read 2 more answers

\int (x+1)\sqrt(2x-1)dx

Nezavi [6.7K]

Answer:

$\int (x+ 1) \sqrt{2x-1} dx = \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15}(2x-1)^{\frac{5}{2}} + C$

Step-by-step explanation:

$\int (x+1)\sqrt {(2x-1)} dx\\Integrate \ using \ integration \ by\ parts \\\\u = x + 1, v'= \sqrt{2x - 1}\\\\v'= \sqrt{2x - 1}\\\\integrate \ both \ sides \\\\\int v'= \int \sqrt{2x- 1}dx\\\\v = \int ( 2x - 1)^{\frac{1}{2} } \ dx\\\\v = \frac{(2x - 1)^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}} \times \frac{1}{2}\\\\v= \frac{(2x - 1)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{2}\\\\v = \frac{2 \times (2x - 1)^{\frac{3}{2}}}{3} \times \frac{1}{2}\\\\v = \frac{(2x - 1)^{\frac{3}{2}}}{3}$

$\int (x+1)\sqrt(2x-1)dx\\\\ = uv - \int v du$

$= (x +1 ) \cdot \frac{(2x - 1)^{\frac{3}{2}}}{3} - \int \frac{(2x - 1)^{\frac{3}{2}}}{3} dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ u = x + 1 => du = dx \ ]$

$= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \int (2x - 1)^{\frac{3}{2}}} dx\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{5}{2}}}{\frac{5}{2} }) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15} \times (2x-1)^{\frac{5}{2}} + C\\\\$

6 0

3 years ago