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3 years ago

What is the side length of a cube that has a total surface area of 96 square centimeters?

Mathematics

2 answers:

Setler79 [48]3 years ago

8 0

96/4. You should get the answer by dividing 96 by 4.

Vedmedyk [2.9K]3 years ago

4 0

The answer is 16
96/6=16

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3 years ago

Solve the quadratic equation. -7n^2=-448

Zolol [24]

Answer:

n = 8

Step-by-step explanation:

-7n² = -448

=> n² = -448 / -7

=> n² = 64

=> n = $\sqrt{64}$

=> n = 8

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3 years ago

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3 years ago

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Solve the equation. –4n + 7 + 2n = 1 A. 1 B. 3 C. –3 D. 4

Kisachek [45]

-2n=-6. n=3. So, b.3.

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3 years ago

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Find all values of k, if any, that satisfy the equation.

aleksklad [387]

Answer:

There are two such values of $k$ , $k = 6-4\sqrt{3}$ and $k = 6+4\sqrt{3}.$

Step-by-step explanation:

The question is to find all values of $k$ , if any, that satisfy the following matrix equation.

$\begin{bmatrix}2 & 2 & k\end{bmatrix} \cdot \begin{bmatrix}1 & 1 & 0\\1 & 0 & 3 \\0 & 3 & -1\end{bmatrix} \cdot \begin{bmatrix}2 \\2\\k\\\end{bmatrix} = 0$

Let's multiply the first two matrices. We can do that, since the number of columns in the first matrix equals the number of rows in the second matrix, which means that their product is defined.

$\begin{bmatrix}2 \cdot 1 + 2 \cdot 1 + k \cdot 0 & 2 \cdot 1 + 2 \cdot 0+ k \cdot 3 & 2 \cdot 0 + 2 \cdot 3 + k \cdot (-1)\end{bmatrix} \cdot \begin{bmatrix}2\\2\\k\end{bmatrix}= 0$

Next, we need to solve the matrix equation

$\begin{bmatrix} 4 & 2+3k & 6-k\end{bmatrix} \cdot \begin{bmatrix}2\\2\\k\end{bmatrix}= 0$

Again, the number of columns in the first matrix equals the number of rows in the second matrix, which means that their product is defined and we can multiply them. The result will be $1$ × $1$ matrix, and that will be the dimension of the zero matrix, as well.

$\begin{bmatrix}4 \cdot 2 + (2+3k) \cdot2 + (6+k)\cdot k\end{bmatrix} = 0\\\begin{bmatrix}8 + 4 + 6 \cdotk + 6 \cdot k - k^2 \end{bmatrix} = 0 & \iff 8 + 4 + 6 \cdot k + 6 \cdot k - k^2 = 0$

Now, all we need to do is to solve the quadratic equation

$-k^2 +12k +12 = 0$

By using the well known formula, we obtain

$k_{1/2} = \frac{-12 \pm \sqrt{12^2 - 4 \cdot (-1) \cdot 12}}{-2} = \frac{-12 \pm \sqrt{192}}{-2} = \frac{-12 \pm \sqrt{64 \cdot 3}}{-2} = \frac{-12 \pm 8\sqrt{3}}{-2}$

Therefore, we obtain two values for $\mathbf{k}$ , $k = 6-4\sqrt{3}$ and $k = 6+4\sqrt{3}$ .

6 0

3 years ago