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3 years ago

0.07 x 1.22 = help thanks

2 answers:

8 0

1.22×0.07=8.54 because hoy múltiple 7×2=14,add the 1 to 2,7×2=14+1=15,add the 1 to 1,las 7×1=7+1=8.The two zeros stay the same,8.54 is your answer

Masteriza [31]3 years ago

4 0

0.07 * 1.22 = 0.0854

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Two roots of a third degree polynomial function f(x) are -4 and 4. Which statement describes the number and nature of all

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Answer:

f(x) has three real roots.

Step-by-step explanation:

f(x) has three real roots.

A third degree polynomial function with real coefficients has exactly 3 roots.

Furthermore, all complex roots come in pairs, either there is none, or there are exactly two conjugates, which are in the form e1+i(e2) and e1-i(e2) where e1 and e2 are polynomial expressions.

This also means that a third degree polynomial with real coefficients has at least one real root, or three.

Since we are given two real roots for the given polynomial, the third root cannot be complex alone, so the only possibility is that there are three real roots.

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3 years ago

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4 years ago

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The limitation of the research include:

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A study will investigate the effectiveness of a new early reading program onPrekindergarten students' reading abilities. The samples (n = 30 and

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2 years ago

Rewrite the system of linear equations as a matrix equation AX = B.

iren2701 [21]

Answer:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]$

And the vector of vriables (X) is:

$\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]$

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

$\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

In conclusion, AX=B is:

$\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]$

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4 years ago

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yanalaym [24]

Answer:

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Step-by-step explanation:

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Answer: 144

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3 years ago