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3 years ago

What is the perimeter of a rectangle with a length of 4 1/4 inches and a width of 2 5/6 inches?

2 answers:

6 0

I'm fairly sure it's C - even as a prediction, if you add (4x2) and (2x2) it's 12 - and the closest is thirteen, adding on the fractions.

kicyunya [14]3 years ago

3 0

D
4 1/4 + 4 1/4 + 2 5/6 + 2 5/6 = 8 1/2 + 2 5/6.
8 1/2 + 2 5/6 = 17/2 + 17/6.
17/2 + 17/6 = 51/6 + 17/6.
51/6 + 17/6 = 68/6.
68/6 = 14 1/6th.

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Which of the following is the equation of a line parallel to the line y=-x+1 passing through the point (4,1)

Softa [21]

Y=-x+b

plug in x = 4 and y = 1 to find b

1 = -4 + b and we know that if you add 5 and -4 you get 1. So ...

5 = b so if we plug that in

y= -x + 5

which is the same as y + x = 5, So it would be A

6 0

3 years ago

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

Now let's calculate the area

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;

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

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.

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

6 0

2 years ago

What is the answer 1 plus 1

AnnZ [28]

Answer:

the answer is 4, just kidding it's 2

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Find the derivative of following function.

Aleks04 [339]

Answer:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}$

General Formulas and Concepts:
Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]:
$\displaystyle (cu)' = cu'$

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
$\displaystyle (uv)' = u'v + uv'$

Derivative Rule [Quotient Rule]:
$\displaystyle \bigg( \frac{u}{v} \bigg)' = \frac{vu' - uv'}{v^2}$

Derivative Rule [Chain Rule]:
$\displaystyle [u(v)]' = u'(v)v'$

Step-by-step explanation:

*Note:

Since the answering box is way too small for this problem, there will be limited explanation.

Step 1: Define

Identify.

$\displaystyle y = \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \times \frac{\tan^2 x + 5x}{\csc^2 x + 3}$

Step 2: Differentiate

We can differentiate this function with the use of the given derivative rules and properties.

Applying Product Rule:

$\displaystyle y' = \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) '$

Differentiating the first portion using Quotient Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( \cos^2 x - 3\sqrt{x} + 6 \big)' \big( \sin^2 x + 6 \big) - \big( \sin^2 x + 6 \big)' \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Differentiating the second portion using Quotient Rule again:

$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( \tan^2 x + 5x \big)' \big( \csc^2 x + 3 \big) - \big( \csc^2 x + 3 \big)' \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule again:
$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Substitute in derivatives:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2} \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Simplify:

$\displaystyle y' = \frac{\big( \tan^2 x + 5x \big) \bigg[ \big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - 2 \sin x \cos x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \bigg]}{\big( \sin^2 x + 6 \big)^2 \big( \csc^2 x + 3 \big)} + \frac{\big( \cos^2 x - 3\sqrt{x} +6 \big) \bigg[ \big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) + 2 \csc^2 x \cot x \big( \tan^2 x + 5x \big) \bigg] }{\big( \csc^2 x + 3 \big)^2 \big( \sin^2 x + 6 \big)}$