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3 years ago

Help with this math homework

Mathematics

1 answer:

rjkz [21]3 years ago

4 0

(Please do keep in mind that when you name angles, the first and third letters can be switched. So if you have <COA and I say <AOC, they are the same thing as long as O is the second letter)

2a) Complementary means the sum of the angles is 90°. Since <COA is 90°, <COB and <BOA are complementary.

2b) Supplementary means the sum of the angles is 180°. Since 180° is a straight line, there are three angles including that angle to make 180°. So in addition to <BOC, your answer would be <BOA and <COD.

2c) Vertical angles are across from each other. The angle across from <BOA is <DOE.

3a) 90°. There's a little square there letting you know it's 90°

3b) 51°. Since <CBF and <CBE are complementary, you would subtract 39 from 90.

3c) 17°. Since <DBE and <ABE are complementary, you subtract 73 from 90.

4) Keep in mind that ray BO is a bisector of <AOC, in other words, the ray splits the angle in half. So <AOB is congruent to <BOC. So solve for x:

x+3=2x+11

3=x+11

x=-8

Now plug in x for either angle

8+3= 11

So that means <AOB and <BOC both equal 11°.

To find <AOC, double that.

<AOC equals 22°.

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I need help writing the first 5 terms of the sequence!

Angelina_Jolie [31]

The first five terms of the sequence are 1, 4, 7, 10, 13.

Solution:

Given data:

$a_{1}=1$

$a_{n}=a_{n-1}+3$

General term of the arithmetic sequence.

$a_{n}=a_{n-1}+d$ , where d is the common difference.

d = 3

$a_{n}=a_{n-1}+3$

Put n = 2 in $a_{n}=a_{n-1}+3$ , we get

$a_{2}=a_1+3$

$a_{2}=1+3$

$a_2=4$

Put n = 3 in $a_{n}=a_{n-1}+3$ , we get

$a_{3}=a_2+3$

$a_{3}=4+3$

$a_3=7$

Put n = 4 in $a_{n}=a_{n-1}+3$ , we get

$a_{4}=a_3+3$

$a_{4}=7+3$

$a_4=10$

Put n = 5 in $a_{n}=a_{n-1}+3$ , we get

$a_{5}=a_4+3$

$a_{5}=10+3$

$a_5=13$

The first five terms of the sequence are 1, 4, 7, 10, 13.

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4 years ago

Explain how to multiply the following whole numbers 21 x 14

Lesechka [4]

Answer:

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Step-by-step explanation:

Given

$21\:\times \:14$

Line up the numbers

$\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}$

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

Multiply the bold numbers: 1×4=4

$\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}$

Multiply the bold numbers: 2×4=8

$\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the top number by the bolded digit of the bottom number

$\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the bold numbers: 1×1=1

$\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}$

Multiply the bold numbers: 2×1=2

$\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}$

Add the rows to get the answer. For simplicity, fill in trailing zeros.

$\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}$

adding portion

$\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}$

Add the digits of the right-most column: 4+0=4

$\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}$

Add the digits of the right-most column: 8+1=9

$\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}$

Add the digits of the right-most column: 0+2=2

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Therefore,

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

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3 years ago

Whats the answer? -4(x-2)-5(y+2)+(-3)(8-6b)

SpyIntel [72]

<span>-4(x-2)-5(y+2)+(-3)(8-6b)
= -4x + 8 - 5y - 10 - 24 + 18b
= -4x -5y + 18b - 26

hope it helps</span>

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4 years ago

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Ivenika [448]

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