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Galina-37 [17]
3 years ago
12

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

Mathematics
2 answers:
cestrela7 [59]3 years ago
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

<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
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:
<u>Calculus</u>

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]:

  1. f(x) = cxⁿ
  2. 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 <em>way</em> too small for this problem, there will be limited explanation.

<u>Step 1: Define</u>

<em>Identify.</em>

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

<u>Step 2: Differentiate</u>

We can differentiate this function with the use of the given <em>derivative rules and properties</em>.

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)}

We can rewrite the differential by factoring and common mathematical properties to obtain our final 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}

∴ we have found our derivative.

---

Learn more about derivatives: brainly.com/question/26836290

Learn more about calculus: brainly.com/question/23558817

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Topic: Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

8 0
2 years ago
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How unspecific? What's the cost of one
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