Factor the polynomial completely using the

Please help me thanks :)

Lina20 [59]

Answer:

A

Step-by-step explanation:

To sovle this we must combine the numbers togther and the variables togther.

We have:

8c+6-3c-2

Lets take out the c's and combine them:

8c -3c

=

5c

Now lets take out the numbers:

6-2

=

4

Now lets put these back together and we get the expression:

5c+4

This looks like A.

Hope this helps! :D

8 0

2 years ago

Javier asks his mother how old a tree in their yard is. His mother says, "The sum of 10 and two-thirds of that tree's age, in ye

Sloan [31]

Answer:

His error is adding 10 and 2/3 before multiplying by a(age of the tree)

Step-by-step explanation:

The sum of 10 and two-thirds of that tree's age, in years, is equal to 50.

Correct equation

Sum = addition (+)

two-thirds = 2/3

The tree's age = a

10 + 2/3a = 50

2/3a = 50 - 10

2/3a = 40

a = 40 ÷ 2/3

= 40 × 3/2

= 60

a = 60 years

Javier writes the equation

(10 + two-thirds) a = 50

(10 + 2/3)a = 50

(30+2/3)a = 50

32/3a = 50

a = 50 ÷ 32/3

= 50 × 3/32

= 150/32

a = 150/32

His error is adding 10 and 2/3 before multiplying by a(age of the tree)

5 0

3 years ago

The equation of function h is h... PLEASE HELP MATH

Flura [38]

Answer:

Part A: the value of h(4) - m(16) is -4

Part B: The y-intercepts are 4 units apart

Part C: m(x) can not exceed h(x) for any value of x

Step-by-step explanation:

Let us use the table to find the function m(x)

There is a constant difference between each two consecutive values of x and also in y, then the table represents a linear function

The form of the linear function is m(x) = a x + b, where

a is the slope of the function
b is the y-intercept

The slope = Δm(x)/Δx

∵ At x = 8, m(x) = 2

∵ At x = 10, m(x) = 3

∴ The slope = $\frac{3-2}{10-8}=\frac{1}{2}$

∴ a = $\frac{1}{2}$

- Substitute it in the form of the function

∴ m(x) = $\frac{1}{2}$ x + b

- To find b substitute x and m(x) in the function by (8 , 2)

∵ 2 = $\frac{1}{2}$ (8) + b

∴ 2 = 4 + b

- Subtract 4 from both sides

∴ -2 = b

∴ m(x) = $\frac{1}{2}$ x - 2

Now let us answer the questions

Part A:

∵ h(x) = $\frac{1}{2}$ (x - 2)²

∴ h(4) = $\frac{1}{2}$ (4 - 2)²

∴ h(4) = $\frac{1}{2}$ (2)²

∴ h(4) = $\frac{1}{2}$ (4)

∴ h(4) = 2

∵ m(x) = $\frac{1}{2}$ x - 2

∴ m(16) = $\frac{1}{2}$ (16) - 2

∴ m(16) = 8 - 2

∴ m(16) = 6

- Find now h(4) - m(16)

∵ h(4) - m(16) = 2 - 6

∴ h(4) - m(16) = -4

Part B:

The y-intercept is the value of h(x) at x = 0

∵ h(x) = $\frac{1}{2}$ (x - 2)²

∵ x = 0

∴ h(0) = $\frac{1}{2}$ (0 - 2)²

∴ h(0) = $\frac{1}{2}$ (-2)² =

∴ h(0) = 2

∴ The y-intercept of h(x) is 2

∵ m(x) = $\frac{1}{2}$ x - 2

∵ x = 0

∴ m(0) = $\frac{1}{2}$ (0) - 2 = 0 - 2

∴ m(0) = -2

∴ The y-intercept of m(x) is -2

- Find the distance between y = 2 and y = -2

∴ The difference between the y-intercepts of the graphs = 2 - (-2)

∴ The difference between the y-intercepts of the graphs = 4

∴ The y-intercepts are 4 units apart

Part C:

The minimum/maximum point of a quadratic function f(x) = a(x - h) + k is point (h , k)

Compare this form with the form of h(x)

∵ h = 2 and k = 0

∴ The minimum point of the graph of h(x) is (2 , 0)

∵ k is the minimum value of f(x)

∴ 0 is the minimum value of h(x)

∴ The domain of h(x) is all real numbers

∴ The range of h(x) is h(x) ≥ 2

∵ m(8) = 2

∵ m(14) = 5

∵ h(8) = $\frac{1}{2}$ (8 - 2)² = 18

∵ h(14) = $\frac{1}{2}$ (14 - 2)² = 72

∴ h(x) is always > m(x)

∴ m(x) can not exceed h(x) for any value of x

Look to the attached graph for more understand

The blue graph represents h(x)

The green graph represents m(x)

The blue graph is above the green graph for all values of x, then there is no value of x make m(x) exceeds h(x)

7 0

3 years ago

R leaks out of a barrel so that the rate of change in the water level is proportional to the square root of the depth of the wat

Irina-Kira [14]

Let y(t) represent the level of water in inches at time t in hours. Then we are given ...
y'(t) = k√(y(t)) . . . . for some proportionality constant k
y(0) = 30
y(1) = 29

We observe that a function of the form
y(t) = a(t - b)²
will have a derivative that is proportional to y:
y'(t) = 2a(t -b)

We can find the constants "a" and "b" from the given boundary conditions.
At t=0
30 = a(0 -b)²
a = 30/b²
At t=1
29 = a(1 - b)² . . . . . . . . . substitute for t
29 = 30(1 - b)²/b² . . . . . substitute for a
29/30 = (1/b -1)² . . . . . . divide by 30
1 -√(29/30) = 1/b . . . . . . square root, then add 1 (positive root yields extraneous solution)
b = 30 +√870 . . . . . . . . simplify

The value of b is the time it takes for the height of water in the tank to become 0. It is 30+√870 hours ≈ 59 hours 29 minutes 45 seconds

5 0

3 years ago

Please someone help me!!! its urgent!

I am Lyosha [343]

Answer:

12.99

Step-by-step explanation:

a²+b²=c²

a²+b²=c²√91²+√78²=c²

a²+b²=c²√91²+√78²=c²√8281+√6084=c²

a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²

a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²169=c²

a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²169=c²c=√169

a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²169=c²c=√169 =13

8 0

2 years ago