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2 years ago

Identify a pattern in the given list of numbers. Then use this pattern to find the next number. 17,7,-3,-13,-23

Mathematics

1 answer:

notka56 [123]2 years ago

4 0

Answer:

-33

Step-by-step explanation:

The sequence is descending so the nth term would be -10n and the 0 term would be 27 so the nth term for the sequence would be -10n +27.

The question asks you to find the 6th term so (-10 x 6) + 27 = -60 + 27 = -33

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How do I solve 2m + n = 2 and 3m - 2n = 3 using substitution?

Leona [35]

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

Solve 2m + n = 2 and 3m - 2n = 3 using substitution.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

We can use the substitution method to solve linear equations of this form. Let's solve for m & n.

$\left. \begin{array} { l } { 2 m + n = 2 } \\ { 3 m - 2 n = 3 } \end{array} \right.$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$2m+n=2, \: 3m-2n=3$

Choose one of the equations and solve it for m by isolating m on the left-hand side of the equal sign.

$2m+n=2$

Subtract n from both sides of the equation.

$2m=-n+2$

Divide both sides by 2.

$m=\frac{1}{2}\left(-n+2\right) \\$

Multiply 1/2 times -n+2.

$m=-\frac{1}{2}n+1 \\$

Substitute $-\frac{n}{2}+1\\$ for m in the other equation, 3m-2n=3.

$3\left(-\frac{1}{2}n+1\right)-2n=3 \\$

Multiply 3 times $-\frac{n}{2}+1\\$ .

$-\frac{3}{2}n+3-2n=3 \\$

Add $-\frac{3n}{2}\\$ to -2n.

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

Subtract 3 from both sides of the equation.

$-\frac{7}{2}n=0 \\$

Divide both sides of the equation by $-\frac{7}{2}\\$ , which is the same as multiplying both sides by the reciprocal of the fraction.

$\large \underline{\underline{ \bf \: n=0 }}$

Substitute 0 for n in $m=-\frac{1}{2}n+1\\$ . Because the resulting equation contains only one variable, you can solve for m directly.

$\large \underline{ \underline{\bf \: m=1 }}$

The system is now solved.

$\huge \boxed{ \boxed{ \bf \: m=1, \: n=0 }}$

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