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Leni [432]
3 years ago
14

Travis reads 21/4 pages of his history assignment in 7 1/2

Mathematics
2 answers:
Bingel [31]3 years ago
5 0

Answer:

B and D

Step-by-step explanation:

Zigmanuir [339]3 years ago
3 0

Answer:

its B and E i just did the question and got it right

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Which relation is not a function?
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It would be w beacuse xy and w
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Write the trinomial as a square of a binomial or as an expression opposite to a square
timurjin [86]
(3a-7)^2

1. 9a^2-21a-21a+49
2. 3a(3a-7)-7(3a-7)
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3 0
2 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}}

7 0
4 years ago
What is the length of VW?
Fudgin [204]
The answer is D. If you make a right triangle with the line, then you can use the Pythagorean theorem (a^2+b^2=c^2) to solve for VW. The vertical line would be 4 units and horizontal line would be 2 units. Therefore 4^2+2^2=20 take the square root of 20 and that is the answer.
5 0
3 years ago
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