Which transformation would not create an image and a pre-image that are CONGRUENT figures?

What is the benchmark fraction of 7/10 x 16

Tomtit [17]

Lets change "16" by its to 16/1 .. since nothing can't help support 16 by itself when it is going into 7/10

Now that we change that ... your problem becomes 7/10 × 16/1

7/10 × 16/1 = 11 1/5

Because
7(16) / 10(1) = 112/10
Now lets go. to how we solve the for the answer

112/10 = 11 1/5

☺

6 0

3 years ago

Pre cal/cal master needed

weqwewe [10]

Take x-2 and insert it into 2x^2 + 3x-2 where the x is located
2x^2 + 3x-2
2(x-2)^2 + 3(x-2)-2

Now work out 2(x-2)^2 + 3(x-2)-2 also follow PEMDAS
2(x-2)^2 + 3(x-2)-2
Since (x-2)^2 is an Exponent, lets work with that first and expand (x-2)^2.
(x-2)^2
(x -2)(x-2)
x^2 -4x + 4
Now Multiply that by 2 because we have that in 2(x-2)^2
(x-2)^2 = x^2 -4x + 4
2(x-2)^2 = 2(x^2 -4x + 4)
2(x^2 -4x + 4) = 2x^2 - 8x + 8
2x^2 - 8x + 8

Now that 2(x-2)^2 is done lets move on to 3(x-2).

Use the distributive property and distribute the 3
3(x-2) = 3x - 6

All that is left is the -2

Now lets put it all together
2(x-2)^2 + 3(x-2)-2

2x^2 - 8x + 8 + 3x - 6 - 2

Now combine all our like terms
2x^2 - 8x + 8 + 3x - 6 - 2
Combine: 2x^2 = 2x^2
Combine: -8x + 3x = -5x
Combine: 8 - 6 - 2 = 0

So all we have left is
2x^2 - 5x

5 0

3 years ago

Solve the equation cos (x/2) = cos x + 1. what are the solutions on the interval 0° ≤ x < 360°?

Sedbober [7]

Answer:

Step-by-step explanation:

cos (x/2)=cos x+1

cos (x/2)=2cos ²(x/2)

2 cos²(x/2)-cos (x/2)=0

cos (x/2)[2 cos (x/2)-1]=0

cos (x/2)=0=cos π/2,cos (3π/2)=cos (2nπ+π/2),cos(2nπ+3π/2)

x/2=2nπ+π/2,2nπ+3π/2