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3 years ago

Plz help me !!!!!!!!

Mathematics

1 answer:

Dovator [93]3 years ago

7 0

Answer: c) -1372

Step-by-step explanation:

$\sqrt[3]{2x} +6=-8\\\\\sqrt[3]{2x} =-14\\\\(\sqrt[3]{2x})^3 =(-14)^3\\\\2x=-2744\\\\\boxed{x=-1372}$

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In an installment loan, a lender loans a borrower a principal amount P, on which the borrower will pay a yearly interest rate of i (as a fraction, e.g. a rate of 6% would correspond to i=0.06) for n years. The borrower pays a fixed amount M to the lender q times per year. At the end of the n years, the last payment by the borrower pays off the loan.

After k payments, the amount A still owed is

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3 years ago

Solve the system of equations using substitution. y = x + 6 y = –2x – 3 (4, –11) (1, 7) (–3, 3)

sergejj [24]

The answer to the question

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3 years ago

Solve the system by using a matrix equation (Picture provided)

Katyanochek1 [597]

Answer:

Option b is correct (8,13).

Step-by-step explanation:

7x - 4y = 4

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it can be represented in matrix form as $\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]$

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X= $\left[\begin{array}{c}x\\y\end{array}\right]$

B= $\left[\begin{array}{c}4\\2\end{array}\right]$

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or X= A⁻¹ B

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= (-42)-(-40)

= (-42) + 40 = -2

so, |A| = -2

Adj A= $\left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]$

A⁻¹ = $\left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]$ / -2

A⁻¹ = $\left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right]$

X= A⁻¹ B

X= $\left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right] *\left[\begin{array}{c}4\\2\end{array}\right]$

X= $\left[\begin{array}{c}(3*4) + (-2*2)\\(5*4) + (-7/2*2)\end{array}\right]$

X= $\left[\begin{array}{c}12-4\\20-7\end{array}\right]$

X= $\left[\begin{array}{c}8\\13\end{array}\right]$

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solution set= (8,13).

Option b is correct.

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3 years ago

How to you put y=(x-8)2+6 in number sense

SpyIntel [72]

Answer:

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Step-by-step explanation:

just that is the way

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IgorC [24]

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3 years ago