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3 years ago

Please help due tonight

Mathematics

1 answer:

AlexFokin [52]3 years ago

5 0

Answer:

hftgftghgdfhgsdjthfdhrtfgredfgesdf

Step-by-step explanation:

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12What mistake did the student make when solvingtheir two-step equation?(a)b) If correctly solved what should the value of be?

mixas84 [53]

Given the equation:

$\frac{x}{6}+3=-18$

(a) You can identify that the student applied the Subtraction Property of Equality by subtraction 3 from both sides of the equation:

$\frac{x}{6}+3-(3)=-18-(3)$

However, the student made a mistake when adding the numbers on the right side.

Since you have two numbers with the same sign on the right side of the equation, you must add them, not subtract them and use the same sign in the result. Then, the steps to add them are:

- Add their Absolute values (their values without the negative sign).

- Write the sum with the negative sign.

Then:

$\frac{x}{6}=-21$

(b) The correct procedure is:

1. Apply the Subtraction Property of Equality by subtracting 3 from both sides (as you did in the previous part):

$\begin{gathered} \frac{x}{6}+3-(3)=-18-(3) \\ \\ \frac{x}{6}=-21 \end{gathered}$

2. Apply the Multiplication Property of Equality by multiplying both sides of the equation by 6:

$\begin{gathered} (6)(\frac{x}{6})=(-21)(6) \\ \\ x=-126 \end{gathered}$

Hence, the answers are:

(a) The student made a mistake by adding the numbers -18 and -3:

$-18-3=-15\text{ (False)}$

(b) The value of "x" should be:

$x=-126$

4 0

1 year ago

Which of the following is the product of the rational expressions shown here? 1/x+2 * x/x-2 A.) x/x^2-4 B.) x+1/x^2-4 C.) x+1/2x

Inessa [10]

Answer:

Step-by-step explanation:

$\frac{1}{x+2}*\frac{x}{x-2}=\frac{1*x}{(x+2)(x-2)}\\\\\\=\frac{x}{x^{2}-2^{2}}\\\\\\=\frac{x}{x^{2}-4}$

5 0

3 years ago

Find the values of λ for which the determinant is zero.

elena55 [62]

Answer:

Determinant are special number that can only be defined for square matrices.

Step-by-step explanation:

Determinant are particularly important for analysis. The inverse of a matrix exist, if the determinant is not equal to zero.

How to find determinant

For a 2×2 matrix

$det ( \left[\begin{array}{cc}x&y\\a&z\end{array}\right] ) = xz-ay$

For a 3×3 matrix

we first decompose it to 2×2

$det (\left[\begin{array}{ccc}k&l&m\\o&p&q\\r&s&t\end{array}\right] )\\\\= k*det(\left[\begin{array}{cc}p&q\\s&t\end{array}\right] ) - l*det(\left[\begin{array}{cc}o&q\\r&t\end{array}\right] ) + m*det(\left[\begin{array}{cc}o&p\\r&s\end{array}\right] ) \\\\=k(pt-sq) - l(ot-rq) + m(os-rp)$

Example

Find the values of λ for which the determinant is zero

$\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right]$

$det(\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right])\\\\= s*det(\left[\begin{array}{cc}s&-1\\-1&1\end{array}\right] ) - (-1)*det(\left[\begin{array}{cc}-1&-1\\0&1\end{array}\right] ) + 0*det(\left[\begin{array}{cc}-1&s\\0&-1\end{array}\right] )\\\\= s(s(1)-(-1*-1)) - (-1)(-1*1 - (-1*0)) + 0\\= s(s - 1)) + 1(-1 + 0) \\=s^{2} -s-1$

Equating the determinant to zero

$s^{2} -s-1 =0\\$

s = $\frac{1}{2}$ * (1 ±5 )

s = 1.61 or -0.61

5 0

3 years ago

Which of the following ratios is NOT equivalent to start fraction 4 over 8 end fraction ?

Novosadov [1.4K]

C

4/8 equals 1/2 the rest are 1/2 but 12/36 is 1/3

6 0

3 years ago

If john had 500 bottles of water and sally took 60263 how much does john have

lukranit [14]

Answer:

John technically doesn't have anymore. Instead, sally owes John 59763 bottles of water. If you're asking for something mathematical, John now has -59763 bottles of water.

4 0

2 years ago

Read 2 more answers