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Fed [463]
4 years ago
13

The midpoint of a line segment is the point halfway between its endpoints. True or False?

Mathematics
2 answers:
MrRa [10]4 years ago
6 0
True because MIDpoint is in the MIDdle which is halfway between two endpoints.

horrorfan [7]4 years ago
5 0
This is true. You can remember by it saying mid, as in middle.
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(18 months). hi I'm gothy and I just needed help with this problem. you don't have to include the answer but I'd appreciate it​
natulia [17]

Answer:

  • $93.68

Step-by-step explanation:

<u>Price of computer:</u>

  • $1486.25

<u>Warranty </u>

  • $199.99

<u>Total</u>

  • $1486.25 + $199.99 = $1686.24

<u>Payment per month:</u>

  • $1686.24/ 18 = $93.68

8 0
3 years ago
Please helppppppp meeeeeeeeeeee
Paha777 [63]

Answer:

the answer is the third one

Step-by-step explanation:

3 0
3 years ago
let a = (a1, a2) and b = (b1, b2) and c = (c1,c2) be three non zero vectors. if a1b2 - a2b1 is not equal to 0. then show three a
Ksenya-84 [330]

Consider the contrapositive of the statement you want to prove.

The contrapositive of the logical statement

<em>p</em> ⇒ <em>q</em>

is

¬<em>q</em> ⇒ ¬<em>p</em>

In this case, the contrapositive claims that

"If there are no scalars <em>α</em> and <em>β</em> such that <em>c</em> = <em>α</em><em>a</em> + <em>β</em><em>b</em>, then <em>a₁b₂</em> - <em>a₂b₁</em> = 0."

The first equation is captured by a system of linear equations,

\begin{cases}c_1 = \alpha a_1 + \beta b_1\\ c_2 = \alpha a_2 + \beta b_2\end{cases}

or in matrix form,

\begin{pmatrix}c_1\\c_2\end{pmatrix} = \begin{pmatrix}a_1&b_1\\a_2&b_2\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}

If this system has no solution, then the coefficient matrix on the right side must be singular and its determinant would be

\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix} = a_1b_2-a_2b_1 = 0

and this is what we wanted to prove. QED

3 0
3 years ago
Which of the following expressions is equal to tan205°?
disa [49]
Tan205° = .466 tan55° = 1.428 tan25° = .466 -tan25° = " so just tan25
8 0
3 years ago
Show the working out too please in numbers, thank you.
vazorg [7]

Hii :))

\tt \: - 5 =  \frac{x}{6}  \\   \tt \:  - 5 \times 6 = x \\  \boxed{  \tt \:  - 30 = x}

__________________

  • The correct value of x is <u>-</u><u> </u><u>3</u><u>0</u><u>.</u>

\overbrace{ \underbrace{ \sf \: \infty  \:  Carry \: on \: Learning \:  \infty }}

ᴛʜᴇᴇxᴛʀᴀᴛᴇʀᴇꜱᴛʀɪᴀʟ

3 0
3 years ago
Read 2 more answers
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