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3 years ago

HELP ASAP PLEASE find the surface area of this triangular prism

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

5 0

Answer:

The answer is 180

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=5x%5E%7B2%7D%2B%7Cx%2B1%7C%5C%20%5Ctextgreater%20%5C%200" id="TexFormula1" title="5x^{2}+|x+1|

Pavel [41]

5x² + |x+1| > 0
5x² + x + 1 > 0
x+1 > 0 because it came from absolute
5x² > 0 because it’s squared

When you try a number you should put it in the original equality
5x² + |x+1|

7 0

2 years ago

( 3 x 2 ) ^ 4 x 3 ^ -3 / 2 ^ 3 x 3 =<br><br> Simplify the exponential expression

Korvikt [17]

Answer:

Step-by-step explanation:

$\frac{(3*2)^4*3^{-3}}{2^3*3}$

we start simplifying by removing the parenthesis

Multiply the exponents inside the the parenthesis

3^4 * 2^4

$\frac{3^4*2^4*3^{-3}}{2^3*3}$

Now we apply exponential property

a^m * a^n = a^ (m+n)

3^4 * 3^-3 = 3^ (4-3) = 3^1

3 or 3^1 are same

$\frac{3^1*2^4}{2^3*3^1}$

3^1 at the top and bottom so we cancel it out

\frac{2^4}{2^3}

we apply log property . a^m / a^n = a^m-n

Now subtract the exponents

2^(4-3) = 2^1 = 2

4 0

4 years ago

Use the following matrices, A, B, C and D to perform each operation.

Vinvika [58]

Step-by-step explanation:

$A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]$

$B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]$

$C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]$

$D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]$

$1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]$

$2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]$

$3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]$

$4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]$