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bekas [8.4K]
3 years ago
10

Quadrilateral ABCD undergoes a reflection across the x-axis to form quadrilateral A'B'C'D'. The coordinates of A' are ( , ).

Mathematics
2 answers:
natka813 [3]3 years ago
5 0
1) A(-5,-3); B(-6,-1); C(-3,-1) ; D(-2,-3)

When a refection is done about x-axis, the values of the abscise x remain identical & the value of ordinate  just change their signs:
A(-5,-3); A'(-5,+3)
B(-6,-1); B'(-6,+1)
C(-3,-1); C'(-3,+1)
D(-2,-3); D'(-2,+3)

2) A'B'C'D' is translated 3 UNITS RIGHT, that means the ordinated of A'B'C'D'
are the same but the abscises have been increased by 3 UNITS ;

A'(-5,+3) ==>A"(-2,3)
B'(-6,+1) ==>B"(-3,1)
C'(-3,+1) ==>C"(0,1)
D'(-2,+3) ==>D"(1,3)
ioda3 years ago
4 0

Answer:

im confused ash

Step-by-step explanation:

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3 years ago
What is the remainder when the polynomial 6x2+11x−3 is divided by 2x−1?
o-na [289]

Answer: 4

There are three different ways to find the remainder.  Since I don't know which lesson you are working on, I will show you all three methods.

<u>Long Division:</u>

           <u>3x  +  7      </u>

2x - 1 ) 6x² + 11x - 3

       -  <u>(6x²  -  3x)</u>   ↓

                    14x  - 3

                -   (<u>14x  - 7)</u>  

                              4    

<u>Synthetic Division:</u>

2x - 1 = 0   ⇒   x = \frac{1}{2}


\frac{1}{2}   | 6      11     -3

    <u>| ↓       3     7</u>

      6      14    4

<u>Remainder Theorem:</u>

2x - 1 = 0   ⇒   x = \frac{1}{2}

f(x) = 6x² + 11x - 3

f(\frac{1}{2}) = 6(\frac{1}{2})² + 11(\frac{1}{2}) - 3

      = 6(\frac{1}{4}) + \frac{11}{2} - 3

      = \frac{3}{2} + \frac{11}{2} - \frac{6}{2}

      = \frac{8}{2}

      = 4

                                     

3 0
3 years ago
Read 2 more answers
Combine like terms in (3x+5)+(4x-30
tresset_1 [31]

Answer:

The answer is 7x-25

Step-by-step explanation:

(3x+5)+(4x-30)

=3x+4x+5-30

=7x-25

3 0
3 years ago
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The function f(x) = 3,267(1 + 0.02)* represents the amount of money in a savings account where x represents time in years. What
andreyandreev [35.5K]

Answer:

C .The initial amount of money placed in the savings account

Step-by-step explanation:

f(x) = 3,267(1 + 0.02)^x

This is in the form

y = a b^x

where a is the initial amount

b is the growth rate

x is the time

3267 is the initial amount

1.02 is the growth rate, so it grows by .02 or 2 percent

and x is the time

6 0
3 years ago
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A ½in diameter rod of 5in length is being considered as part of a mechanical linkagein which it can experience a tensile loading
Evgesh-ka [11]

Answer:

a. Maximum Load = Force = 27085.09 N

b. Maximum Energy = 3.440 Joules

Step-by-step explanation:

Given

Rod diameter = ½in = 0.5in

Length = 5in

Young's modulus = 15.5Msi

By applying the 0.2% offset rule,

The maximum load the rod can hold before it gets to breaking point is given as follows by taking the strain as 0.2%

Young Modulus = Stress/Strain ------- Make Stress the Subject of Formula

Stress = Strain * Young Modulus

Stress = 0.2% * 15.5

Stress = 0.002 * 15.5

Stress = 0.031Msi

Calculating the area of the rod

Area = πr² or πd²/4

Area, A = 22/7 * 0.5^4 / 4

A = 22/7 * 0.25 / 4

A = 5.5/28

A = 0.1964in²

The maximum load that the rod would take before it starts to permanently elongate is given by

Force = Stress * Atea

Force = 0.031Msi * 0.1964in²

Force = 31Ksi * 0.1964in²

Force = 6.089Ksi in²

Force = 6.089 * 1000lbf

Force = 6089 lbf

1 lbf = 4.4482N

So, Force = 6089 * 4.4482N

Force = 27085.09 N

b.

Using Strain to Energy Formula

U = V×σ²/2·E

Where V = Volume

V = Length * Area

V = 5 in * 0.1964in²

V = 0.982in³

σ = Stress = 0.031Msi

= 0.031 * 1000Ksi

= 31Ksi

= 31 * 1000psi

= 31000psi

E = Young Modulus = 15.5Msi

= 15.5 * 1000Ksi

= 15.5 * 1000 * 1000psi

= 15500000psi

So,

Energy = 0.982 * 31000²/ ( 2 * 15500000)

Energy = 943,702,000/31000000

Energy = 30.442in³psi

------- Converted to ftlbf

Energy = 2.537 ftlbf

-------- Converted to Joules

Energy = 3.440 Joules

7 0
4 years ago
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