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nadezda [96]
3 years ago
12

Find the length of a segment with an endpoint of (2, 1) and a midpoint of (–2, –1).

Mathematics
1 answer:
Travka [436]3 years ago
5 0

Answer:

Length of segment =4\sqrt5 units = 8.94 units

Step-by-step explanation:

Given:

End point of line: (2,1)

Midpoint of line: (-2,-1)

To find length of line segment.

Solution:

Distance from endpoint to midpoint is half the length of segment as the midpoint divides the line into equal halves.

Distance from end point to mid point can be found out using distance formula:

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

Plugging in points for end point (2,1) and midpoint(-2,-1).

d=\sqrt{(-2-2)^2+(-1-1)^2)

d=\sqrt{(-4)^2+(-2)^2)

d=\sqrt{16+4)

d=\sqrt{20}

d=2\sqrt{5}

Length of half segment =2\sqrt{5} units

∴ Length of segment = 2\times length\ of\ half\ segment= 2\times 2\sqrt5=4\sqrt5 units

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Solve |3k-2|=2|k+12|
Stella [2.4K]

Answer:

k = 26   or   k = -\dfrac{22}{5}

Step-by-step explanation:

|3k - 2| = 2|k + 12|

\dfrac{|3k - 2|}{|k + 12|} = 2

|\dfrac{3k - 2}{k + 12}| = 2

3k - 2 = 2(k + 12)   or   3k - 2 = -2(k + 12)

3k - 2 = 2k + 24   or   3k - 2 = -2k - 24

k = 26   or   5k = -22

k = 26   or   k = -\dfrac{22}{5}

7 0
3 years ago
The model represents x2 – 9x + 14. An algebra tile configuration showing only the Product spot. 24 tiles are in the Product spot
klio [65]

I didn't get all the part with the tiles, but here's the general answer:

given a polynomial

p(x)=ax^2+bx+c

we have that x-k is a factor of p(x) if and only if k is a root of p(x), i.e. if

p(k)=ak^2+bk+c=0

So, given the polynomial

p(x)=x^2-9x+14

We can check if x-9 is a factor by evaluating p(9):

p(9)=81-81+14=14\neq 0

So, x-9 is not a factor.

Similarly, we can evaluate p(2),\ p(-5),\ p(-7) to check if x-2,\ x+5,\ x+7 are factors:

p(2)=4-18+14=0,\quad p(-5)=25+45+14=84\neq 0,\quad p(-7)=49+63+14=126 \neq 0

So, only x-2 is a factor of x^2-9x+14

4 0
3 years ago
Read 2 more answers
Does the following equation determine y to be a function of x? y square=x+3
Phantasy [73]

Answer:

∴ y² = x + 3 is not a function

Step-by-step explanation:

* Lets explain how to solve the problem

- The definition of the function is every input (x) has only one

  output (y)

- Ex:

# y = x + 1 where x ∈ R , is a function because every x has only

  one value of y

# y² = x where x ∈ R , is not a function because y = ±√x, then one

  x has two values of y

* Lets solve the problem

∵ y² = x + 3

- Find y by taking √ for both sides

∴ y = ± √(x + 3)

- That means y = √(x + 3)  and y = - √(x + 3)

∵ (x + 3) must be greater than or equal zero because there is no

  square root for negative number

∴ x + 3 ≥ 0 ⇒ subtract 3 from both sides

∴ x ≥ -3

∴ x must be any number greater than or equal -3

- Let x = 0

∴ y = √(0 + 3) = √3 and y = - √(0 + 3) = -√3

∴ x = 0 has two values of y ⇒ y = √3 and y = -√3

- Any value of x greater than or equal 3 will have two values of y

∴ y² = x + 3 is not a function

5 0
3 years ago
Factor and simplify 8cos^2x+cosx-9
KonstantinChe [14]
HELLO DEAR,




GIVEN:-
8cos²x + cosx - 9

8cos²x + 9cosx - 8cosx - 9

cosx(8cosx + 9) - (8cosx + 9)

(cosx - 1)(8cosx + 9)


I HOPE ITS HELP YOU DEAR,
THANKS
3 0
3 years ago
Lisa invested $5200 in a savings account with a yearly interest rate of 6% for
andrey2020 [161]

P is the principal amount, $5200.00.

r is the interest rate, 6% per year, or in decimal form, 6/100=0.06.

t is the time involved, 9....year(s) time periods.

So, t is 9....year time periods.

To find the simple interest, we multiply 5200 × 0.06 × 9 to get that: 

The interest is: $2808.00

4 0
3 years ago
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