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joja [24]
3 years ago
15

PLEASE HELP ASAP!!!!

Mathematics
1 answer:
My name is Ann [436]3 years ago
8 0

Answer:

CHEATER. EXAM QUESTION

Step-by-step explanation:

YA NAIA GUY. I AM MORE NAIA THAN NAIA

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Name two properties used to evaluate 7x1-4x1/4
Kitty [74]
7x1=7, then 4x1/4= 4x1=4 divided by 4= 1. Then subtract 7- 1= 6
5 0
3 years ago
Use a transformation to solve the equation. w/4 = 8 can you also leave a detailed explanation on how this equation = 32
OverLord2011 [107]

Answer:

w=32

Step-by-step explanation:

w/4 = 8

4/1 * w/4 = 8*4

w=32

5 0
3 years ago
Answer these questions
seropon [69]
You should divide 43 from 126
3 0
3 years ago
The points -5,2 and -5,8 lie on a circle with a radius of 3. Find the center of the circle.
worty [1.4K]

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

We have two points A(-5, 2) and B(-5, 8). Substitute:

d=\sqrt{(8-2)^2+(-5-(-5))^2}=\sqrt{6^2+0^2}=\sqrt{6^2}=6

d is a diameter of a circle because d = 2r = 2(3).

The center of a circle is in midpoint of AB.

The formula of a midpoint:

M\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)

Substitute:

\dfrac{-5+(-5)}{2}=\dfrac{-10}{2}=-5\\\\\dfrac{2+8}{2}=\dfrac{10}{2}=5

Answer: (-5, 5).

6 0
3 years ago
a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point.
xxMikexx [17]

Answer:

y = 13*( -x/9 + 1/5)

Step-by-step explanation:

Given:

- The curve has an equation as follows:

                               44 = 5x^2 + 3xy + 3y^2

Find:

a. Verify that the given point (2​,2​) lies on the curve.

b. Determine an equation of the line tangent to the curve at the given point.

Solution:

- To verify whether the point lies on the given curve we will substitute the coordinates of the point into the equation as follows:

                              44 = 5*(2)^2 + 3*(2)(2) + 3*(2)^2

                              44 = 20 + 12 + 12

                              44 = 44 ......Hence proven.

- The equation of the line tangent to the curve is expressed as a linear function as follows:

                              y = m*x + C

Where, m is the gradient of the line.

            C is the y-intercept.

                              m = Δy / Δx = dy/dx

- We will take the derivative of the given curve with respect to x as follows:

                             0 = 10x + 3*( y + xy' )  + 6y*y'\\\\-10x - 3y = y' ( 3x + 6y)\\\\ y' = - \frac{10x + 3y}{3x + 6y}

- Evaluate y' at the point (2,2) we get:

                            y' = - ( 10(2) + 3(2) ) / ( 3(2) + 6(2) )

                            y' = - ( 26 ) / (18)

                            y'= m = - 13/9

- To evaluate C, we will use the point (2,2) for linear expression above with m as follows:

                            y = -13*x/9 + C

                            2 =-13*(2)/9 + C

                            C = 13 / 5

- The equation of the tangent is as follows:

                            y = 13*( -x/9 + 1/5)  

8 0
3 years ago
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