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3 years ago

How are patterns used to solve problems?

Mathematics

2 answers:

vladimir1956 [14]3 years ago

8 0

Yes you can use patterens

aleksandrvk [35]3 years ago

7 0

Your can use patterns in line graphs, bar chats, pie charts, tables.
Patterns themselves can help solve maths problems, eg;
2,4,6,8,10.
It has a pattern of add two every time.

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Evaluate 17 + 70/x when x = 14.

kipiarov [429]

Answer:

34.5

Step-by-step explanation:

17 + 70/x when x = 14.

substitute x for 14

17 + 70/4

17 + 17.5

34.5

5 0

3 years ago

Pls help me on this question :-

Nastasia [14]

Answer:

We conclude that the rule for the table in terms of x and y is:

y = 3x+2

Step-by-step explanation:

The table indicates that there is constant change in the x and y values, meaning the table represents the linear function the graph of which would be a straight line.

We know the slope-intercept form of the line equation

y = mx+b

where m is the slope and b is the y-intercept.

Taking two points

(-2, -4)

(-1, -1)

Finding the slope between (-2, -4) and (-1, -1)

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(-2,\:-4\right),\:\left(x_2,\:y_2\right)=\left(-1,\:-1\right)$

$m=\frac{-1-\left(-4\right)}{-1-\left(-2\right)}$

$m=3$

We know that the y-intercept can be determined by setting x = 0 and finding the corresponding y-value.

Taking another point (0, 2) from the table.

It means at x = 0, y = 2.

Thus, the y-intercept b = 2

Using the slope-intercept form of the linear line function

y = mx+b

substituting m = 3 and b = 2

y = 3x+2

Therefore, we conclude that the rule for the table in terms of x and y is:

y = 3x+2

7 0

3 years ago

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

4 0

3 years ago

Please help!!! On my hw

neonofarm [45]

Answer:

c(X)=R(X)

125x+18000=215x

18000=215x-125x

18000=90x (divede both sides by 90)

X=200

3 0

3 years ago

If you say that you can help me with homework then why can't I make a code to join and I had to pick one to send this sincerely,

Nataly_w [17]

Answer:

WHAT!!!!!!!

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers