When a depependent variable increases and the independent variable increases what is the constant rate of Change

1 answer:

4 0

Maybe In a linear equation, the independent variable increases at a constant rate while the dependent variable decreases at a constant rate.

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You found one formula, area of △ABC=(1/2)as sin B, that applies to an obtuse triangle just as it does to an acute triangle. Whic

ira [324]

Let ΔABC be an obtuse triangle with sides a, b, c, that lie opposite the angles A, B, C; with radius r of inscribed circle and radius R of circumscribed circle; p semiperimeter. For obtuse triangle you can use all known area formulas:

1. $A= \frac{1}{2} ac\sin B$ ; (the area is equal to half of product of two sides and sine of angle that lies between these sides);

2. $A= \frac{1}{2} ah_a$ ;
3. $A= \sqrt{p(p-a)(p-b)(p-c)}$ (Heron's formula);
4. $A=pr$ ;
5. $A= \dfrac{abc}{4R}$ .

7 0

3 years ago

Which expression is equivalent to (3x^2-7)?

Musya8 [376]

Answer:

The correct answer is third option

(10x² - 4) - (7x² + 3)

Step-by-step explanation:

From the given option we get the correct answer is third option.

(10x² - 4) - (7x² + 3)

<u>Check the correct answer</u>

(10x² - 4) - (7x² + 3) = 10x² - 4 - 7x² - 3)

= 10x² - 7x² - 4 - 3 = 3x² - 7

Therefore the correct answer is third option

(10x² - 4) - (7x² + 3)

4 0

3 years ago

Read 2 more answers

Match each situation to its corresponding equation

Nitella [24]

First pair goes to f(x) = 20x
Next pair goes to f(x) = 10x + 20
Third pair goes to f(x) = 10x + 25
Last pair goes to f(x) = 20x + 10

7 0

4 years ago

If the endpoints of the diameter of a circle are (6, 3) and (2, 1), what is the standard form equation of the circle?

saul85 [17]

(x - 4 )² + ( y - 2 )² = 5 → d

the equation of a circle in standard form is

(x - a )² + ( y - b)² = r²

where (a , b) are the coordinates of the centre and r is the radius

the centre is at the midpoint of the endpoints of the diameter and the radius is the distance from the centre to either of the 2 endpoints of the diameter.

to find the centre use the midpoint formula

(a , b) = [ $\frac{1}{2}$ (6 + 2) , $\frac{1}{2}$ (3 + 1 )] = (4 , 2)