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3 years ago

!PLEASE HELP WILL GIVE BRAINLIEST!

Mathematics

2 answers:

rewona [7]3 years ago

7 0

Answer:

$let \: cost \: be \: { \bf{c}} \: and \: months \: be \: { \bf{n}} \\ { \bf{c \: \alpha \: n}} \\ { \bf{c = kn}} \\ 34 = (k \times 1) \\ k = 34 \: dollars \\ \\ { \boxed{ \bf{c = 34n}}}$

natita [175]3 years ago

4 0

Answer:

34m = c

Step-by-step explanation:

For every month (m) you pay 34 dollars. However many months youu use that service time 34 equals your total cost (c).

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What two numbers multiply to get 1440 but add to get 78

Maslowich

The answer is 68, Which should be correct

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3 years ago

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17 points)Which figures are similar to the one without a letter please explain I’m a little confused.....?plz help

erik [133]

Answer:

The letter D

Step-by-step explanation:

because of the present in the top-left corner has no letter and it maches with the present that has no color and that is the letter D XD thanks brainleiiest plz

8 0

3 years ago

Find the value of w-9 given that –3w+5 = 2

melamori03 [73]

Answer:

$-8$

Step-by-step explanation:

Solving for $w$ given the equation, $-3w +5 = 2$ :

$-3w +5 = 2 \\ -3w +5 -5 = 2 -5 \\ -3w = -3 \\ \frac{-3w}{-3} = \frac{-3}{-3} \\ w = 1$

Solving for $w -9$ when $w = 1$

$1 -9 \\ -8$

3 0

2 years ago

23 x 32 is the prime factorization for which one of these choices?

antoniya [11.8K]

<h2>Answer:</h2>

⟶ 2³ × 3² is the prime factorization for which one of these choices?

Let's check,

1) 6 = 3 × 2 [So, obviously not this choice]

2) 25 = 5 × 5 = 5² [Not this either]

3) 36 = 3 × 2 × 2 × 3 = 3² × 2² [Doesn't match with 2³ × 3²]

4) 72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3² [Matches]

⟶ The answer is, choice 72.

$\underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}}$

3 0

2 years ago

Find the determinant of the nxn matrix A with 5's on the diagonal, l's above the diagonal, and O's below the diagonal. det (A) =

Vilka [71]

Answer:

n times 5

Step-by-step explanation:

A matrix Anxn of this way is called an upper triangular matrix. It can be proved that the determinant of this kind of matrix is

$a_{11}+a_{22}+...+a_{nn}$

In this case, it would be 5+5+...+5 (n times) = n times 5

We are going to develop each determinant by the first column taking as pivot points the elements of the diagonal

$det\left[\begin{array}{cccc}5&a_{12}&a_{13}...&a_{1n}\\0&5&a_{23}...&a_{2n}\\...&...&...&...\\0&0&0&5\end{array}\right] =5+det\left[\begin{array}{ccc}5&a_{23}...&a_{2n}\\0&5&a_{3n}\\...&...&...\\0&0&5\end{array}\right]=5+5+...+det\left[\begin{array}{cc}5&a_{n-1,n}\\0&5\end{array}\right]=5+5+...+5+5\;(n\;times)$

6 0

3 years ago