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3 years ago

Please help asap! it’s area and perimeter of rectangles and parallelograms!

Mathematics

1 answer:

Rama09 [41]3 years ago

7 0

Answer: I don’t know how to solve the area of these shapes. But I know how to find the peeimeter Of this shape.

perimeter=P

finding the P of any shape is like finding the P of a square or rectangle. Is to add up all the sides up like for the 8th one it would be 10+20+20+10 or (10x2)+(20x2)

hope this helps!

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B2x%2B4%7D%7Bx%5E%7B2%7D-2x-8%20%7D" id="TexFormula1" title="\frac{2x+4}{x^{2}-2x-8 }

lisabon 2012 [21]

Answer:

$\frac{2}{x-4}$

Step-by-step explanation:

$\frac{2x+4}{x^2-2x-8}$

Factor both the numerator and denominator

$\frac{2(x+2)}{(x-4)(x+2)}$

The $x+2$ 's cancel out

The equation in its simplest form is $\frac{2}{x-4}$

Hope this helps!

6 0

3 years ago

Which of these constructions is impossible using only a compass and straightedge

mariarad [96]

Answer:

neko...

Step-by-step explanation:

4 0

3 years ago

Solve 4(3x+3)=15x+7-3x+5

Flura [38]

Answer:

Infinite amount of solutions

Step-by-step explanation:

Step 1: Write equation

4(3x + 3) = 15x + 7 - 3x + 5

Step 2: Solve for x

Distribute 4: 12x + 12 = 15x + 7 - 3x + 5
Combine like terms: 12x + 12 = 12x + 12
Subtract 12 on both sides: 12x = 12x
Divide both sides by 12: x = x

Here, we can see that x would be infinite amount of solutions. We can plug in any number x and it would render the equation true.

4 0

3 years ago

Jake is solving x^2-6x+5=0 by completing the square his steps are shown below step 1. x^2-6x=-5 step 2. x^-6x+9=-5+9 step 3. (x-

damaskus [11]

Answer:

Jake's error in step 3

Step-by-step explanation:

we have

$x^{2} -6x+5=0$

Complete the square

step 1

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2} -6x=-5$

step 2

Complete the square. Remember to balance the equation by adding the same constants to each side

$x^{2} -6x+9=-5+9$

$x^{2} -6x+9=4$

step 3

Rewrite as perfect squares

$(x-3)^{2}=4$

Jake's error in step 3

He placed 6 instead of 3 in the left side

step 4

take square root both sides

$\sqrt{(x-3)^2} =\sqrt{4}$

step 5

$(x-3)=\pm2$

step 6

$x=3\pm2$

step 7

x=5 and x=1

7 0

4 years ago

8+9x1x2 to the second power it has the 2 above the 2 but I just can't put it there on this

Gnoma [55]

The correct answer is 44

8 0

3 years ago