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insens350 [35]
3 years ago
12

Find the equation in the line shown please Is for today

Mathematics
1 answer:
Olegator [25]3 years ago
3 0

Step-by-step explanation:

i hope i have been useful buddy.

good luck ♥️♥️♥️♥️♥️.

You might be interested in
1. Write an algebra equation for each purchase. I (Mickey) bought an ice cream treat and a bottle of water for $10. Pam purchase
Ne4ueva [31]

x+y=10

x+2y=14
x is common in both so u have to subtract the second one from 1st
x-x + y-2y = 10-14
-y = -4
y=4
which means bottle of water (y) = 4
and then substitute 4 in equation (x+y=10 or x+2y=10)
x + 4 = 10
x=6
5 0
3 years ago
Questions 16-17 | Math 1 - 0 points Solve the graph Help needed !!
iris [78.8K]

Answer:

16) The area of the circle is 25.1 units²

17) JKLM is a parallelogram but not a rectangle

Step-by-step explanation:

16) Lets talk about the area of the circle

- To find the area of the circle you must find the length of the radius

- In the problem you have the center of the circle and a point on

 the circle, so you can find the length of the radius by using the

 distance rule

* Lets solve the problem

∵ The center of the circle is (1 , 3)

∵ The point on the circle is (3 , 5)

- Using the rule of the distance between two points

* Lets revise it

- The distance between the two points (x1 , y1) and (x2 , y2) is:

 Distance = √[(x2 - x1)² + (y2 - y1)²]

∴ r = √[(3 - 1)² + (5 - 3)²] = √[4 + 4] = √8 = 2√2 units

∵ The area of the circle = πr²

∴ The area of the circle = π (2√2)² = 8π = 25.1 units²

* The area of the circle is 25.1 units²

17) To prove a quadrilateral is a parallelogram, prove that every

     to sides are parallel or equal or the two diagonal bisect

     each other

* The parallelogram can be rectangle if two adjacent sides are

  perpendicular to each other (measure of angle between them is 90°)

 or its diagonals are equal in length

- The parallel lines have equal slopes, then to prove the

  quadrilateral is a parallelogram, we will find the slopes of

  each opposite sides

* Lets find from the graph the vertices of the quadrilateral

∵ J = (0 ,2) , K (2 , 5) , L (5 , 0) , M (3 , -3)

- The opposite sides are JK , ML and JM , KL

- The slope of any line passing through point (x1 , y1) and (x2 , y2) is

 m = (y2 - y1)/(x2 - x1)

∵ The slop of JK = (5 - 2)/(2 - 0) = 3/2 ⇒ (1)

∵ The slope of LM = (-3 - 0)/(3 - 5)= -3/-2 = 3/2 ⇒ (2)

- From (1) and (2)

∴ JK // LM

∵ The slope of KL = (0 - 5)/(5 - 2) = -5/3 ⇒ (3)

∵ The slope of JM = (-3 - 2)/(3 - 0)= -5/3 ⇒ (4)

- From (3) and (4)

∴ KL // JM

∵ Each two opposite sides are parallel in the quadrilateral JKLM

∴ It is a parallelogram

- The product of the slopes of the perpendicular line is -1

* lets check the slopes of two adjacent sides in the JKLM

∵ The slope of JK = 3/2 and the slope of KL = -5/3

∵ 3/2 × -5/3 = -5/2 ≠ -1

∴ JKLM is a parallelogram but not a rectangle

4 0
3 years ago
You will get a loan from a bank for $5000. The interest rate will be 8% and the loan will be for 3 years. You will need to calcu
Leviafan [203]

Answer:

5000×8×3/100

1200

1200×10/100

120

3 0
2 years ago
Please can someone answer this
Alecsey [184]

Answer:

theres not alot of info it just says boys, and ends there

Step-by-step explanation:

4 0
3 years ago
The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s
oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is A(6) = 28.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable t.

Notice that the interval of interest 0 \le t \le 6 is closed on both ends. In other words, this interval includes both endpoints: t = 0 and t= 6. Over this interval, the value of A(t) might be maximized when t is at the following:

  • One of the two endpoints of this interval, where t = 0 or t = 6.
  • A local maximum of A(t), where A^\prime(t) = 0 (first derivative of A(t)\! is zero) and A^{\prime\prime}(t) (second derivative of \! A(t) is smaller than zero.)

Start by calculating the value of A(t) at the two endpoints:

  • A(0) = 10.
  • A(6) = 28.

Apply the power rule to find the first and second derivatives of A(t):

\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}.

\displaystyle A^{\prime\prime}(t) = 6\, t - 15.

Notice that both t = 1 and t = 4 are first derivatives of A^{\prime}(t) over the interval 0 \le t \le 6.

However, among these two zeros, only t = 1\! ensures that the second derivative A^{\prime\prime}(t) is smaller than zero (that is: A^{\prime\prime}(1) < 0.) If the second derivative A^{\prime\prime}(t)\! is non-negative, that zero of A^{\prime}(t) would either be an inflection point (ifA^{\prime\prime}(t) = 0) or a local minimum (if A^{\prime\prime}(t) > 0.)

Therefore \! t = 1 would be the only local maximum over the interval 0 \le t \le 6\!.

Calculate the value of A(t) at this local maximum:

  • A(1) = 15.5.

Compare these three possible maximum values of A(t) over the interval 0 \le t \le 6. Apparently, t = 6 would maximize the value of A(t)\!. That is: A(6) = 28 gives the maximum value of \! A(t) over the interval 0 \le t \le 6\!.

However, note that the maximum over this interval exists because t = 6\! is indeed part of the 0 \le t \le 6 interval. For example, the same A(t) would have no maximum over the interval 0 \le t < 6 (which does not include t = 6.)

4 0
3 years ago
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