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2 years ago

Name all the properties of a square

Mathematics

1 answer:

vovikov84 [41]2 years ago

7 0

Answer:

Here :)

Equilateral Polygon,

Convex Polygon,

Cyclic

Isgonal Figure

Istoxal Figure

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If m∠ACD = 136° and m∠B = 42°, Find m∠A.

miss Akunina [59]

Answer:

2°

Step-by-step explanation:

Hello

I see you are very confused by this problem. However, there is an easy way to solve this one.

It has been proved that the sum of all angles in a rectangle is always 180°. Here is a way to prove it:

x y

_______A_________

/ \

___/_______ \__

B C

we have two parellel lines xy and BC. It is true that:

xAy = 180°

xAB + BAC + CAy = xAy = 180

xAB = ABC

CAy= ACB

=> ........

I'll let you finish with your great intellect ^_^

HOPE YOU LEARN WITH JOY AND GOOD GRADE

3 0

2 years ago

5th Grade U.S Studies Weekly- Week 24 Answer Key

Georgia [21]

You should try and put it in another way it doesn’t make sense

3 0

2 years ago

Can anyone help me with this please and ty I really appreciate it if u can

Anarel [89]

Your riiiiiiiiiiuiuuiii

8 0

2 years ago

Find the missing side

dybincka [34]

Answer:

7) 8.94

8) 6.71

Step-by-step explanation:

for both questions, use pythagorean theorem

$a^{2} + b^{2} = c^{2}$

Question 7:

$4^{2} + 8^{2} = c^{2}$

16 + 64 = $c^{2}$

80 = $c^{2}$

$\sqrt{80} = \sqrt{c^{2}}$

c = 8.94

Question 8:

$6^{2} + 3^{2} = c^{2}$

36 + 9 = $c^{2}$

45 = $c^{2}$

$\sqrt{45} = \sqrt{c^{2} }$

c = 6.71

6 0

2 years ago

For 20 Points. 3 Qs. I also gave you guys a help sheet. PLEASE HELP.

Eduardwww [97]

Wat is help sheet, didn't help,

instructions unclear, got my calculator stuck in the ceiling fan

remember some simple exponential laws
$( x^{n} )(x^{m})=x^{n+m}$ and
$\frac{x^{n}}{x^{m}} =x^{ n-m }$
$x^{-m}= \frac{1}{x^{m}}$

first one
$(2^{-6})(2^{3})=2^{-6+3}=2^{-3}= \frac{1}{2^{3}} = \frac{1}{8}$

2nd one
$\frac{7^{5}}{7^{3}} =7^{ 5-3 } = 7^{2} = 49$

third one
$(3^{-4})(3^{2})=3^{-4+2}=3^{-2}= \frac{1}{3^{2}} = \frac{1}{9}$

dunno wat to do with 'help sheet'

4 0

2 years ago