What is the unit rate?

On the interval [x,7] , at which x-value is the average rate of change 56?<br> ????????

Ainat [17]

Answer:

4

Step-by-step explanation:

i got it right on the test

3 0

3 years ago

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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

Find the discriminant to determine the number of roots, then solve the quadratic equation using the quadratic formula. !! PLS HE

VikaD [51]

Answer:

Discriminant review

Discriminant reviewThe discriminant is the part of the quadratic formula underneath the square root symbol: b²-4ac. The discriminant tells us whether there are two solutions, one solution, or no solutions.

7 0

3 years ago

Whats the reciprocal of 9/14

velikii [3]

14/9 reciprocal is a reverse of the number so it should be opposite
i hope it helps

4 0

3 years ago

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First 100 units = 24p per unit Remaining units = 16p per unit Mrs. Watt checks her electricity bill. Here are her meter readings

Talja [164]

Answer: a) 146 units used b) the cost of the number of units used is 3136 p.

Step-by-step explanation:

Since we have given that

First 100 units = 24 p per unit

Remaining units = 16 p per unit

New reading = 7143 units

Old reading = 6997 units

Difference between the readings would be

$7143-6997\\\\=146$

So, there are 146 units used.

(B) Calculate the cost of the number of units used

There are 146 units

For the first 100 units = 24 p per unit

For the remaining i.e. 146-1400 = 46 units , cost = 16 p per unit

Total cost of the number of units used is given by

$24\times 100+46\times 16\\\\=2400+736\\\\=3136\ p$

Hence, a) 146 units used b) the cost of the number of units used is 3136 p.

7 0

3 years ago