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WITCHER [35]
3 years ago
9

Is 7 to the 8th power times 7 equivalent to 7 to the the third power times 7 to third power equivalent?

Mathematics
1 answer:
Akimi4 [234]3 years ago
8 0

Answer:

They are not equivalent.

Step-by-step explanation:

Before we solve the question, we need to know that from the laws of indices:

A^x x A^y = A^(x+y).

7 to the power of 8th power means 7^8

7^8 times 7 means 7^8 x 7. From the laws of indices we have 7^8 x 7 = 7^9.

7 to the power of 3rd power means 7^3

7^3 times 7 to the power of 3rd power means 7^3 x 7^3. From the laws of indices we have 7^3 x 7^3 = 7^6.

Obviously, 7^9 ≠ 7^3 therefore, they are not equivalent.

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In the figure, the slope of mid-segment DE is -0.4. The slope of segment AC is?
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Answer:

\pink{\boxed{{\colorbox{white}{Answer}}}}

\purple {\overline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}

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A.) 0.4

B.) -0.4

C.) 2

D.) -2

\purple {\overline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}

Step-by-step explanation:

\pink{\boxed{{\colorbox{white}{- MsTiyana}}}}

# i hope it helps you

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8 0
2 years ago
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If we have two similar triangles:
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3 years ago
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3 years ago
The matrix C=[1,-2_-3,7] was used to encode a phrase to [7,-28,-25,-35,-2_-21,107,90,123,17]. Find C^-1 and use it to decode the
posledela
C=  \left[\begin{array}{cc}1&-2\\-3&7\end{array}\right]  \\ C^{-1}= \left[\begin{array}{cc}7&2\\3&1\end{array}\right]  \\ \left[\begin{array}{cc}7&2\\3&1\end{array}\right] \times   \left[\begin{array}{ccccc}7&-28&-25&-35&-2\\-21&107&90&123&17\end{array}\right]  \\ =\left[\begin{array}{ccccc}49-42&-196+214&-175+180&-245+246&-14+34\\21-21&-84+107&-75+90&-105+123&-6+17\end{array}\right] \\ =\left[\begin{array}{ccccc}7&18&5&1&20\\0&23&15&18&11\end{array}\right]
is the required matrix.
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3 years ago
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