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3 years ago

Can someone please help me with 15?

Mathematics

1 answer:

Feliz [49]3 years ago

4 0

the answer is m=17

https://photomath.net/s/argDe9 shows how to get there

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Solve for n<br> 2/5n+3-7/10n=3/5-3n

LuckyWell [14K]

<h2>The answer is " $-\frac{8}{9} \\\\$ ".</h2>

Given:

$\to \frac{2}{5}n+3-\frac{7}{10}n=\frac{3}{5}-3n$

To find:

n=?

Solution:

$\to \frac{2}{5}n+3-\frac{7}{10}n=\frac{3}{5}-3n\\\\\to \frac{2}{5}n-\frac{7}{10}n+3n=\frac{3}{5} -3\\\\\to n(\frac{4-7+30}{10})=\frac{3-15}{5} \\\\\to n(\frac{-3+30}{10})=\frac{3-15}{5} \\\\\to n(\frac{27}{10})=\frac{-12}{5} \\\\\to n=\frac{-12}{5} \times \frac{10}{27} \\\\\to n=-4 \times \frac{2}{9} \\\\\to n=-\frac{8}{9} \\\\$

Therefore, the final answer of n is " $-\frac{8}{9} \\\\$ ".

Find out more information about solving here:

brainly.com/question/322601

7 0

3 years ago

Put the quadratic into vertex form and state the coordinates of the vertex. y= y= \,\,x^2-6x-3 x 2 −6x−3

enot [183]

Answer:

$y = (x - 3)^2 - 12$

$(3,-12)$

Step-by-step explanation:

Given

$y = x^2 - 6x - 3$

Solving (a): In vertex form

The vertex form of an equation is:

$y = a(x - h)^2 + k$

To do this, we make use of completing the square method.

We have:

$y = x^2 - 6x - 3$

------------------------------------------------------------------

Take the coefficient of x (i.e. -6)

Divide by 2; -6/2 = -3

Square it: (-3)^2 = 9

Add and subtract the result to the equation

------------------------------------------------------------------

$y = x^2 - 6x - 3$

$y = x^2 - 6x + 9 - 9 - 3$

$y = x^2 - 6x + 9 - 12$

Factorize $x^2 - 6x + 9$

$y = x^2 - 3x-3x + 9 - 12$

$y = x(x - 3)-3(x - 3) - 12$

Factor out x - 3

$y = (x - 3)(x - 3) - 12$

Express as squares

$y = (x - 3)^2 - 12$

Hence, the vertex form of $y = x^2 - 6x - 3$ is: $y = (x - 3)^2 - 12$

Solving (b): State the coordinates of the vertex.

In $y = a(x - h)^2 + k$ ; the vertex is: (h,k)

The vertex of $y = (x - 3)^2 - 12$ will be $(3,-12)$

3 0

3 years ago

What is the solution to | 2x-4| >8?

Serga [27]

$| 2x-4| \ \textgreater \ 8 \\ \\ 2x - 4 \ \textgreater \ 8 \ , \ -(2x - 4) \ \textgreater \ 8 \ / \ break \ problem \ into \ 2 \ equations \\ \\ first \ equation \\ \\ 2x \ \textgreater \ 8 + 4 \ / \ add \ 4 \ to \ each \ side \\ \\ 2x \ \textgreater \ 12 \ / \ simplify \\ \\ x \ \textgreater \ \frac{12}{2} \ / \ divide\ each \ side \ by \ 2 \\ \\ x \ \textgreater \ 6 \ / \ simplify \\ \\ second \ equation \\ \\ -2x + 4 \ \textgreater \ 8 \ / \ simplify \\ \\ -2x \ \textgreater \ 8 - 4 \ / \ subtract \ 4 \ from \ each \ side \\ \\ -2x \ \textgreater \ 4 \ / \ simplify \\ \\ x \ \textless \ \frac{4}{-2} \ / \ divide \ each \ side \ by \ -2 \\ \\$
$\\ \\ x \ \textless \ - \frac{4}{2} \ / \ simplify \\ \\ x \ \textless \ - 2 \ / \ simplify \\ \\ collect \ solutions \\ \\ x \ \textgreater \ 6 \ , \ x \ \textless \ -2 \\ \\$

So, your answer to this problem is x > 6 and x < -2.

8 0

3 years ago

If sin just answer the please lol i don't feel like typing<br><br> ASAP

yaroslaw [1]

Considering that the sine is negative and that the cosine is positive, the angle is on the fourth quadrant, hence option C is correct.

<h3>What are the signs of the sine and of the cosine in each quadrant?</h3>

Quadrant 1: Both positive.

Quadrant 2: Sine positive, cosine negative.

Quadrant 3: Both negative.

Quadrant 4: Sine negative, cosine positive.

Hence, since the sine is negative and that the cosine is positive, the angle is on the fourth quadrant, hence option C is correct.

More can be learned about quadrants at brainly.com/question/28021191

#SPJ1

5 0

2 years ago

Divisor of number <br>6?

mash [69]

Answer:

1 2 3 6

Step-by-step explanation:

1x6=6

2x3=6

3x2=6

6x1=6

6 0

3 years ago