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3 years ago

What do you already know about linear relationships?

Mathematics

1 answer:

inna [77]3 years ago

8 0

Answer:

A linear relationship (or linear association) is a statistical term used to describe a straight-line relationship between two variables. Linear relationships can be expressed either in a graphical format or as a mathematical equation of the form y = mx + b.

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Question 2(Multiple Choice Worth 5 points) (05.01)A poster is shown below: A rectangle is shown. The length of the rectangle is

Karo-lina-s [1.5K]

Answer:

The enlargement will be 30 ft by 7.5 ft.

Step-by-step explanation:

A scale drawing is a larger or smaller version of an object. To find the dimensions of the new drawing, multiply each measurement by the scale factor. The scale factor here is 5/2 or 2.5.

12(2.5)=30 ft

3(2.5)=7.5 ft

7 0

2 years ago

Pls answer ASAP<br> 5 extra points

Leona [35]

Answer:

The answer is B.

Step-by-step explanation:

Any percentage is a number divided by 100 . So 0.4%=0.4100 . Dividing 0.4 by 100 gives 0.004 .

5 0

3 years ago

An item costs $410 before tax, and the sales tax is $20.50 . find the sales tax rate. write your answer as a percentage.

Blababa [14]

Hi there! For this, we can simply divide 20.50 by 410 to find the tax rate. 20.50/410 is 0.05. That's in decimal form. Multiply by 100 to get the percent form. 0.05 * 100 is 5. That's the number in percent form. The sales tax rate is 5%.

6 0

2 years ago

Read 2 more answers

Find the surface area of the triangular prism. and equation plz

Firdavs [7]

Answer:

648

Step-by-step explanation:

A=Ph

P=perimeter of triangle

P=6+6+6

P=18

h= height of prism

h=38

A=ph

A=18*36

A=648

4 0

2 years ago

Read 2 more answers

Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8

Ivanshal [37]

It looks like the differential equation is

$\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$

Factorize the right side by grouping.

$xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)$

$xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)$

Now we can separate variables as

$\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx$

Integrate both sides.

$\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx$

$\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx$

$\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}$

You could go on to solve for $y$ explicitly as a function of $x$ , but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0

1 year ago