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seraphim [82]
3 years ago
15

Really hard geometry problems. Please help!

Mathematics
2 answers:
zvonat [6]3 years ago
7 0

Answer:

i think #3 is 93 and #4 is 70

Step-by-step explanation:

a full circle is equal to 360

for #3, i used the equation (2x-10) + (x+4) + (x+10) = 360

combine all like terms and get 4x + 4 = 360

subtract 4 from both sides

4x + 4 = 360

4x = 356

divide 4 from both sides

x = 89. from here put 89 where x is . angle BGD and angle CGA are the same so 89 + 4 = 93

for #4, use the equation 45 + 3x - 19 = 110

combine like terms and get 26 + 3x = 110

subtract 26 from both sides and then divide by 3

3x = 84

x = 28

place 28 where x is and find the matching angle which is 70.

HOPE THIS HELPS!!

Bogdan [553]3 years ago
6 0

Answer: For #3 the answer is 48 degrees, #4 is 70 degrees

Step-by-step explanation:

3. CGA and BGD are verticle angles which means they are congruent, and (2x-10)+(x+4)+(x+10) is 180. Which gives you X=44, then you substitute x into x+4 which gives you 48.

4. CEB and BED are supplementary which means they make up 180. BED and CEA are verticle angles which means they are congruent. We know CEB is 110 which means BED is 70 because 180-110=70. BED and CEA are verticle angles which means CEA is also 70 degrees

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Answer:

- 3.5229

Step-by-step explanation:

Using the rules of logarithms

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log_{10} 10^{n} = n

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log_{10} 0.0003

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= log_{10} 3 + log_{10} 10^{-4}

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Step-by-step explanation:

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Read 2 more answers
Which expression represents the number 2i4−5i3+3i2+−81‾‾‾‾√ rewritten in a+bi form?
vichka [17]

Answer:

The expression -1+14i represents  the number 2i^4-5i^3+3i^2+\sqrt{-81} rewritten in a+bi form.

Step-by-step explanation:

The value of i is i=\sqrt{-1}[tex] or [tex]i^{2}=-1[\tex].Now [tex]i^{4} in term of i^{2}[\tex] can be written as, [tex]i^{4}=i^{2}\times i^{2}

Substituting the value,

i^{4}=\left(-1\right)\times \left(-1\right)

Product of two negative numbers is always positive.

\therefore i^{4}=1

Now i^{3} in term of i^{2}[\tex] can be written as, [tex]i^{3}=i^{2}\times i

Substituting the value,

i^{3}=\left(-1\right)\times i

Product of one negative  and one positive numbers is always negative.

\therefore i^{3}=-i

Now \sqrt{-81} can be written as follows,

\sqrt{-81}=\sqrt{\left(81\right)\times\left(-1\right)}

Applying radical multiplication rule,

\sqrt{ab}={\sqrt{a}}\sqrt{b}

\sqrt{\left(81\right)\times\left(-1\right)}={\sqrt{81}}\sqrt{-1}

Now, \sqrt{\left(81\right)=9 and \sqrt{-1}}=i

\therefore \sqrt{\left(81\right)\times\left(-1\right)}=9i

Now substituting the above values in given expression,

2i^4-5i^3+3i^2+\sqrt{-81}=2\left(1\right)-5\left(-i\right)+3\left(-1\right)+9i

Simplifying,

2+5i-3+9i

Collecting similar terms,

2-3+5i+9i

Combining similar terms,

-1+14i

The above expression is in the form of a+bi which is the required expression.

Hence, option number 4 is correct.

5 0
3 years ago
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