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77julia77 [94]
3 years ago
5

May someone help me with 2-9? Please?

Mathematics
1 answer:
mylen [45]3 years ago
7 0

2. Each side of a pentagon is the same size.

4cm x 5 = 20cm or 4cm+4cm+4cm+4cm+4cm = 20cm

3. Each side of a square is the same size.

13yd x 4 = 52yd or 13yd+13yd+13yd+13yd = 52yd

4. Add all sides together.

12m+12m+30m+30m = 84m

5. Again add all sides together.

16yd+16yd+4yd+4yd = 40yd

6. Each side of a square is the same size.

7in x 4 = 28in. or 7in+7in+7in+7in = 28in

7. Add all sides together.

2cm+2cm+3cm+3cm = 10cm

8. Each side of a rhombus is the same size. A rhombus has 4 sides.

23in x 4 = 92in or 23in+23in+23in+23in = 92in

9. A regular octagon has 8 sides and each side is the same size.

9cm x 8 = 72cm

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FinnZ [79.3K]
1500.*3.75%= 56.25*11= 618.75
3 0
3 years ago
Read 2 more answers
Someone please help me
allsm [11]

Answer:

Solutions: x = \frac{-3}{ 4} + i \sqrt{39},  x = \frac{-3}{4} - i \sqrt{39}

Step-by-step explanation:

Given the quadratic equation, 2x² + 3x + 6 = 0, where a =2, b = 3, and c = 6:

Use the <u>quadratic equation</u> and substitute the values for a, b, and c to solve for the solutions:

x = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}

x = \frac{-3 +/- \sqrt{3^{2} - 4(2)(6)} }{2(2)}

x = \frac{-3 +/- \sqrt{9- 48} }{4}

x = \frac{-3 +/- \sqrt{-39} }{4}

x = \frac{-3 + i \sqrt{39} }{4}, x = \frac{-3 - i \sqrt{39} }{4}

Therefore, the solutions to the given quadratic equation are:

x = -\frac{3}{ 4} + i \sqrt{39} ,   x = -\frac{3}{4} - i \sqrt{39}

4 0
2 years ago
I need help quick thanks!!!
aliya0001 [1]

Answer:

x = 1.5

Step-by-step explanation:

-1.2 (3 - x) = -1.8

<em>Apply distributive property</em>

-3.6 + 1.2x = -1.8

<em>Get 1.2x alone so add -3.6 to each side</em>

1.2x = 1.8

<em>Get x alone so divide 1.2 on each side</em>

x = 1.5

Hope this helps!

4 0
2 years ago
Help?pleaseeee?someone?anyone?
slamgirl [31]

Answer:

the answer is 96

Step-by-step explanation:

(-10)^{2}- 9(\frac{1}{3} )^{2}-3

100-\frac{9}{9}-3

100-1-3

96

6 0
3 years ago
What is the linear function equation represented by the graph?
igor_vitrenko [27]

Look at the picture.

The point-slope form of a line:

y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (0, 1) and (2, 0). Substitute:

m=\dfrac{0-1}{2-0}=\dfrac{-1}{2}=-\dfrac{1}{2}\\\\y-1=-\dfrac{1}{2}(x-0)\\\\y-1=-\dfrac{1}{2}x\qquad|+1\\\\y=-\dfrac{1}{2}x+1

Answer: f(x)=-\dfrac{1}{2}x+1

5 0
3 years ago
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