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Black_prince [1.1K]
3 years ago
11

Name the type of polynomial indicating the

Mathematics
1 answer:
lutik1710 [3]3 years ago
5 0

first off, bear in mind that the  terms are usually separated by a + or a - sign.

secondly, the degree of a term of a polynomial comes from the sum of its variables' exponents.

a)

\bf 2x^3y^2\qquad \begin{cases} \textit{no + or - signs, just one term a \underline{monomial}}\\\\ x^3y^2\implies \stackrel{degree}{3+2}\implies 5 \end{cases}

b)

\bf \stackrel{term}{5x}+\stackrel{term}{3x^2}-\stackrel{term}{9}\qquad \begin{cases} \textit{three terms, a \underline{trinomial}}\\\\ 5x^1\implies 1\\ 3x^2\implies 2\qquad \stackrel{\textit{degree of the polynomial}}{\textit{term with highest degree}}\\ \end{cases}

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Hopefully this answered your question!

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Given the functions flx) = 3x2, g(x) = x2 - 4x + 5, and h(x) = -2x2 + 4x + 1, rank them from leasrto greatest based on their axi
MakcuM [25]

9514 1404 393

Answer:

  f(x), h(x), g(x)

Step-by-step explanation:

For a quadratic in standard form, ...

  ax² +bx +c

the axis of symmetry is ...

  x = -b/(2a)

For the various functions, the axes of symmetry are ...

  f(x): x = 0/(2·3) = 0

  g(x): x = -(-4)/(2·1) = 2

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From least to greatest axis of symmetry the functions are ...

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4 0
3 years ago
Find the area of the triangle with the given vertices. Use the fact that the area of the triangle having u and v as adjacent sid
Gnom [1K]

Answer:

The area of the triangle is A=\sqrt{\frac{4027}{2}}

Step-by-step explanation:

Using the fact that the area of the triangle having u and v as adjacent sides is given by

A=\frac{1}{2}||{\bf u} \times {\bf v} ||

We know that we want to take a cross product to compute the area of the triangle, but we need to be careful because it doesn't make sense if we take the cross product of points.

The first step is to build some vectors that describe this triangle.

According with the graph we can build the vectors:

{\bf AB} and {\bf AC}

The vector {\bf AB} is the difference of point B minus point A

{\bf AB}=(5-3,5-5,0-9)=(2,0,-9)

and the vector {\bf AC} is the difference of point C minus point A

{\bf AC}=(-4-3,0-5,2-9)=(-7,-5,-7)

Next we need to find the cross product of this vectors.

{\bf AB} \times {\bf AC}=\begin{pmatrix}2&0&-9\end{pmatrix}\times \begin{pmatrix}-7&-5&-7\end{pmatrix}

This is the definition of cross product of two vectors in space:

Let {\bf u} = u_1{\bf i}+u_2{\bf j}+u_3{\bf k} and {\bf v} = v_1{\bf i}+v_2{\bf j}+v_3{\bf k} be vectors in space. The cross product of {\bf u} and {\bf v} is the vector

{\bf u} \times {\bf v}=(u_2v_3-u_3v_2){\bf i}-(u_1v_3-u_3v_1){\bf j}+(u_1v_2-u_2v_1){\bf k}

Applying this definition we get

{\bf AB} \times {\bf AC}=\begin{pmatrix}2&0&-9\end{pmatrix}\times \begin{pmatrix}-7&-5&-7\end{pmatrix}

\begin{pmatrix}0\cdot \left(-7\right)-\left(-9\left(-5\right)\right)&-9\left(-7\right)-2\left(-7\right)&2\left(-5\right)-0\cdot \left(-7\right)\end{pmatrix}\\\\\begin{pmatrix}-45&77&-10\end{pmatrix}

||{\bf AB} \times {\bf AC}||=\sqrt{(-45)^2+(77)^2+(-10)^2} \\\\||{\bf AB} \times {\bf AC}||=\sqrt{2025+5929+100}\\\\||{\bf AB} \times {\bf AC}||=\sqrt{8054}

The area of the triangle is

A=\frac{1}{2}||{\bf AB} \times {\bf AC} ||=\frac{1}{2}\sqrt{8054}=\sqrt{\frac{4027}{2}}

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Lapatulllka [165]

Answer:

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If the digit in the ones place is greater than 5, we increase the tens place by 1 & convert the ones place into a 0.

If not, we leave the tens place as is & convert the ones place into a 0.

Since 3 is less than 5, we round 553 to 550 so you are correct.

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