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3 years ago

Without dividing or listing multiples how can you tell if 61 is a multiple of 6.

1 answer:

7 0

For a number to be a multiple of 6, it must be divisible by 6
and a number divisble by 6 means that it is divisble by 2 and 3

since the ones digit is odd it is not divisble by 2 which means 61 is not a multiple of 6

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FIND THE EXACT AREA PLEASE HELPPPP ILL GIVE BRAINLIESTTTT

Ede4ka [16]

Answer:

555.94

Step-by-step explanation:

16 x 11 = 176/2

since there are two of the same triangle

88 x 2

176

11 is the radius

A = $\pi$ $x^{2}$

A = 3.14 ( pie is 3.14, if not then pi) x (11x11)

A = 3.14 x 121 (or 121 $\pi$ )

A = 379.94

379.94 + 176 (or 121pi + 176)

555.94

4 0

2 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

p ⇒ q

¬q ⇒ ¬p

In this case, the contrapositive claims that

"If there are no scalars α and β such that c = αa + βb, then a₁b₂ - a₂b₁ = 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

Three friends, Maria, Sarah, and Ruth, spend $356.50, $365.50, and $356.60,

Likurg_2 [28]

Answer:

Im guessing ruth...

Step-by-step explanation:

not very clear on question

7 0

2 years ago

What is the solution, if any, to the inequality |3x| ≥0?

Yanka [14]

Answer:

the answer is either c or b

4 0

2 years ago

Read 2 more answers

the square root of 4,356 is an integer. without a calculator , determine what the integer is by eliminating possibilities.

Zanzabum

It's more than 3600, so its greater than 60.
it's less than 4900, so its less than 70.
a no.'s root can end with six if the no. ends with 4 or 6.
so the no is either 64 or 66.
Now if it's 66, the square must be greater than sq of 65.
65 sq= 4225
the given no. is greater than sq of 65.
Therefore the number is 66.

5 0

3 years ago