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klemol [59]
3 years ago
10

What is the answer for 23-2|2-4(3-6)|

Mathematics
1 answer:
Alexxandr [17]3 years ago
5 0

use PEMDAS

The numbers in the absolute value symbols are what you need to solve first:

I2-4(3-6)I

I2-4(-3)I                  simplify the parentheses

I2+12I                     multiply -4 and -3

I14I                         add the terms 2 and 12

14                           simplify 14 with the absolute value symbols


Now solve what's on the outside of the absolute value symbols:

23-2

21              subtract and simplify 23 and -2


Finally, multiply the two terms we have made, 14 and 21:

14 * 21 = 294


Your final answer should be 294.

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Solve for x <br> (1 point)
Delicious77 [7]
It should be 13 since 13+8 = 21. 21-7 = 13
5 0
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What are the endpoint coordinates for the midsegment of △PQR that is parallel to PQ¯¯¯¯¯?
andriy [413]

Answer:

M(x₄ ,y₄) = (-3.5 , 0.5)  and

N (x₅ ,y₅) = ( -1 , -0.5 )

Step-by-step explanation:

Let the endpoint coordinates for the mid segment of △PQR that is parallel to PQ be

M(x₄ ,y₄) and N(x₅ ,y₅) such that MN || PQ

point P( x₁ , y₁) ≡ ( -3 ,3 )

point Q( x₂ , y₂) ≡ (2 , 1 )

point R( x₂ , y₂) ≡ (-4 , -2)  

To Find:

M(x₄ ,y₄) = ?  and

N (x₅ ,y₅) = ?

Solution:

We have Mid Point Formula as

Mid\ point(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})

As M is the mid point of PR and N is the mid point of RQ so we will have

Mid\ pointM(x_{4} ,y_{4})=(\frac{x_{1}+x_{3} }{2}, \frac{y_{1}+y_{3} }{2})

Mid\ pointN(x_{5} ,y_{5})=(\frac{x_{2}+x_{3} }{2}, \frac{y_{2}+y_{3} }{2})

Substituting the given value in above equation we get

Mid\ pointM(x_{4} ,y_{4})=(\frac{-3+-4 }{2}, \frac{3+-2} }{2})

∴ Mid\ pointM(x_{4} ,y_{4})=(\frac{-7} }{2}, \frac{1}{2})

∴ Mid\ pointM(x_{4} ,y_{4})=(-3.5, 0.5)

Similarly,

Mid\ pointN(x_{5} ,y_{5})=(\frac{2+-4 }{2}, \frac{1+-2 }{2})

∴ Mid\ pointN(x_{5} ,y_{5})=(\frac{-2 }{2}, \frac{-1}{2})

∴ Mid\ pointN(x_{5} ,y_{5})=(-1, -0.5)

∴ M(x₄ ,y₄) = (-3.5 , 0.5)  and

  N (x₅ ,y₅) = ( -1 , -0.5 )

3 0
3 years ago
Which equivalent expression would you set up to verify the associative property of addition for (3x + 4) + ((5x2 – 1) + (2x + 6)
saw5 [17]

(3x+4)+(5x2-1)+(2x+6)

remove unnecessary parenthesis

3x+4+(5x2-1)+(2x+6)

3x+4+(10-1)+2x+6

subtract the numbers

3x+4+9+2x+6

collect the like terms

5x+4+4+9+6

5x+19

4 0
3 years ago
Read 2 more answers
An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

8 0
3 years ago
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