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polet [3.4K]
3 years ago
11

Help in the question above please

Mathematics
1 answer:
Shkiper50 [21]3 years ago
4 0

Answer:

A)11

Step-by-step explanation:

These are matrices one dimensional with one column and 3 rows each.

-The product of the matrices is obtained by multiplying the correspond values and summing up;

pq=\left[\begin{array}{ccc}3\\2\\-1\end{array}\right] \times\left[\begin{array}{ccc}5\\-1\\2\end{array}\right] \\\\\\\\=(3\times 5)+(2\times -1)+(-1\times 2)\\\\=15+-2+-2\\\\=11

Hence, the product of p and q is 11

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Whats does pemdas mean
Bumek [7]
Its a common technique for remembering the order of operations. The abbreviation pemdas is turned into the phrase "Please excuse my Dear Aunt Sally". It stands for "Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction.
8 0
3 years ago
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A. ASA Postulate <br> B. HL Theorem <br> C.AAS Theorem<br> D. SAS Postulate
TiliK225 [7]

Answer:

your answer will be <em><u>B. HL Theorem </u></em>

Step-by-step explanation:

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3 years ago
How many 1/2 foot pieces can be cut from 6 foot ribbon?
MatroZZZ [7]
Number of ribbons that can be cut = 6 ÷ 1/2 
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Number of ribbons that can be cut = 12 

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Answer: 12  1/2 foot pieces can be cut from 6 feet ribbon
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8 0
3 years ago
The frequency of the musical note C4 is about 261.63 Hz. What is the frequency of the note a perfect fifth below C4?
kolbaska11 [484]

Answer:

The frequency of the note a perfect fifth below C4 is;

B- 174.42 Hz

Step-by-step explanation:

Here we note that to get the "perfect fifth" of a musical note  we have to play a not that is either 1.5 above or 1.5 below the note to which we reference. Therefore to get the frequency of the note a perfect fifth below C4 which is about 261.63 Hz, we have

1.5 × Frequency of note Y = Frequency of C4

1.5 × Y = 261.63

Therefore, Y = 261.63/1.5 = 174.42 Hz.

6 0
3 years ago
A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th
olchik [2.2K]

Answer:

14.52 seconds.

Step-by-step explanation:

We have been given that the height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation y=-16x^2+224x+121. We are asked to find the time, when the rocket will hit the ground.

We know that the rocket will hit the ground, when height will be 0. So to find the time when rocket will hit the ground, we will substitute y=0 in our given equation as:

0=-16x^2+224x+121

Let us solve for x using quadratic formula.

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\frac{-224\pm\sqrt{224^2-4(-16)(121)}}{2(-16)}

x=\frac{-224\pm\sqrt{50176+7744}}{-32}

x=\frac{-224\pm\sqrt{57920}}{-32}

x=\frac{-224\pm240.66574}{-32}

x=\frac{-224+240.66574}{-32}, x=\frac{-224-240.66574}{-32}

x=\frac{16.66574}{-32}, x=\frac{-464.66574}{-32}

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Upon rounding to nearest 100th of second, we will get:

x\approx -0.52, x\approx 14.52

Since time cannot be negative, therefore, the rocket will hit the ground after 14.52 seconds.

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3 years ago
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