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2 years ago

13

Use the clues below to determine a value for point C. Write your answer in the space provided and explain why your answer is cor

rect.

1 answer:

AfilCa [17]2 years ago

4 0

C= -1 because B has to be smaller than -2 so B represents -3 and C has to be less than 0.
The value for C is -1

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Verify that the points are the vertices of a parallelogram and find its area. (2,-1,1), (5, 1,4), (0,1,1), (3,3,4)

Gelneren [198K]

Answer:

Area = 13.15 square units

Step-by-step explanation:

Let the given vertices be represented as follows:

A(2, -1, 1) = 2i - j + k

B(5, 1, 4) = 5i + j + 4k

C(0, 1, 1) = 0i + j + k

D(3, 3, 4) = 3i + 3j + 4k

(i) Let's calculate the vectors of all the sides:

$\\$ AB = B - A = (5i + j + 4k) - (2i - j + k)

AB = 5i + j + 4k - 2i + j - k [Collect like terms]

AB = 3i + 2j + 3k

BC = C - B = (0i + j + k) - (5i + j + 4k)

BC = 0i + j + k - 5i - j - 4k [Collect like terms]

BC = -5i + 0j - 3k

CD = D - C = (3i + 3j + 4k) - (0i + j + k)

CD = 3i + 3j + 4k - 0i - j - k [Collect like terms]

CD = 3i + 2j + 3k

DA = A - D = (2i - j + k) - (3i + 3j + 4k)

DA = 2i - j + k - 3i - 3j - 4k [Collect like terms]

DA = -i - 4j - 3k

AC = C - A = (0i + j + k) - (2i - j + k)

AC = 0i + j + k - 2i + j - k [Collect like terms]

AC = -2i + 2j

BD = D - B = (3i + 3j + 4k) - (5i + j + 4k)

BD = 3i + 3j + 4k - 5i - j - 4k [Collect like terms]

BD = -2i + 2j

(ii) From the results in (i) above, we can deduce that;

AB = CD This implies that AB || CD [AB is parallel to CD]

AC = BD This implies that AC || BD [AC is parallel to BD]

(iii) Therefore, ABDC is a parallelogram since opposite sides (AB and CD) are parallel. Hence, the points are vertices of a parallelogram

<u>Now let's calculate the area</u>

To find the area of the parallelogram, we find the magnitude of the cross product of any two adjacent sides.

In this case, we'll choose AB and AC

Area = |AB X AC|

Where;

$AB X AC = \left[\begin{array}{ccc}i&j&k\\3&2&3\\-2&2&0\end{array}\right]$

<u></u>

AB X AC = i(0 - 6) - j(0 + 6) + k(6 + 4)

AB X AC = - 6i - 6j + 10k

|AB X AC| = $\sqrt{(-6)^2 + (-6)^2 + (10)^2}$

|AB X AC| = $\sqrt{172}$

|AB X AC| = 13.15

Area = 13.15 square units.

<u></u>

<u></u>

<u>PS: </u> ACBD is also a parallelogram. The diagram has also been attached to this response.

6 0

2 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

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3 years ago

Choose the congruence theorem that you would use to prove the triangles congruent.

Elena L [17]

<u><em>Answer:</em></u>

SAS

<u><em>Explanation:</em></u>

<u>Before solving the problem, let's define each of the given theorems:</u>

<u>1- SSS (side-side-side):</u> This theorem is valid when the three sides of the first triangle are congruent to the corresponding three sides in the second triangle

<u>2- SAS (side-angle-side):</u> This theorem is valid when two sides and the included angle between them in the first triangle are congruent to the corresponding two sides and the included angle between them in the second triangle

<u>3- ASA (angle-side-angle):</u> This theorem is valid when two angles and the included side between them in the first triangle are congruent to the corresponding two angles and the included side between them in the second triangle

<u>4- AAS (angle-angle-side):</u> This theorem is valid when two angles and a side that is not included between them in the first triangle are congruent to the corresponding two angles and a side that is not included between them in the second triangle

<u>Now, let's check the given triangles:</u>

We can note that the two sides and the included angle between them in the first triangle are congruent to the corresponding two sides and the included angle between them in the second triangle

This means that the two triangles are congruent by <u>SAS</u> theorem

Hope this helps :)

5 0

3 years ago

Read 2 more answers

Graph the line with slope <br> 2<br> 5<br><br> and <br> y<br><br> -intercept <br> −<br> 7<br><br> .

tia_tia [17]

Answer:

See attachment

Step-by-step explanation:

We want to graph the line with slope

$m = \frac{2}{5}$

and y-intercept ,

$b = - 7$

The equation of this line is

$y = \frac{2}{5}x - 7$

Starting at -7 on the y-axis, we run 5 units and rise 2 units right.

Then draw a line through the y intercept and the new point

The graph is shown in the attachment.

3 0

3 years ago

Plz someone help me with this problem

fiasKO [112]

Not factorable
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6 0

3 years ago