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stiks02 [169]
3 years ago
13

Given that AXB is complementary to both CYD and FZE and m AXB=20 degrees what is m CYD+m FZE?

Mathematics
2 answers:
ohaa [14]3 years ago
8 0
We know that AXB + CYD and AXB + FZE = 90

20 + CYD = 90  and  20 + FZE = 90

We can see that CYD and FZE are equal so we can use one equation

Let's use 20 + CYD = 90

CYD = 70 and FZE is also equal to 70 because  CYD and FZE are equal.

so 70 + 70 = 140 degree

makkiz [27]3 years ago
4 0
It is given that, AXB + CYD = 90
So, CYD = 90 - AXB
CYD = 90 - 20 = 70

Agaian, AXB + FZE = 90
FZE = 90 - AXB
FZE = 90 - 20 = 70

So, CYD + FZE = 70 + 70 = 140

In short, Your Answer would be: 140 degrees

Hope this helps!
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7 0
3 years ago
Simplify the variable expression below as much as possible.<br> 3x + 2y + 5-3
agasfer [191]

Answer:

3x+2y+2=0

3x=-2y-2

x=-(2y+2)/3

Step-by-step explanation:

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3 years ago
879,100 rounded to the nearest ten thousand
faust18 [17]
880,000 is the answer
5 0
3 years ago
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What is the value of A when we rewrite... (PLZ HELP QUICK)
Marina86 [1]

Answer:

<h2>\frac{133}{8}</h2>

Step-by-step explanation:

Given,

{( \frac{5}{2} )}^{x}  +  {( \frac{5}{2} )}^{x + 3}

=  {( \frac{5}{2}) }^{x}  +  {( \frac{5}{2}) }^{x}  \times  {( \frac{5}{2} )}^{3}

= ( \frac{5}{2} ) ^{x} (1 +  {( \frac{5}{2} )}^{3}

=  {( \frac{5}{2} )}^{x} (1 +  \frac{125}{8} )

=  {( \frac{5}{2} )}^{x} ( \frac{1 \times 8 + 125}{8} )

=  {( \frac{5}{2}) }^{x} ( \frac{8 + 125}{8} )

{( \frac{5}{2} )}^{x} ( \frac{133}{8} )

Comparing with A • {( \frac{5}{2}) }^{x}

A = \frac{133}{8}

Hope this helps...

Good luck on your assignment...

7 0
3 years ago
(200 x 3) + (50 x 3)
m_a_m_a [10]

<em><u>200x3=600</u></em>

<em><u>50x3=150</u></em>

<em><u></u></em>

<em><u></u></em>

<em><u>600+150=750</u></em>

<em><u></u></em>

<em><u>750</u></em>

6 0
3 years ago
Read 2 more answers
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