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Stells [14]
2 years ago
12

Which could NOT be side lengths of a parallelogram?

Mathematics
2 answers:
Kipish [7]2 years ago
4 0

Answer:

The answer is c as the sides in a parallelogram need to be equal to the opposite side making them the same length.

Mama L [17]2 years ago
4 0

Answer:C. 4 m, 3 m, 4 m, 4 m

Step-by-step explanation:

A parallelogram is a quadrilateral in which the opposite sides are parallel. The opposite sides and opposite angles are also equal.

Looking at the given options,

A) the opposite sides are equal to 2 yards. Therefore, it is a parallelogram.

B) the opposite sides are equal to 2cm and 3cm each. Therefore, it is a parallelogram.

C)A pair of opposite sides is equal to 4 cm and the other pair of opposite sides is unequal because one side is 3m and the other is 4 m. Therefore, it is not a parallelogram.

D) the opposite sides are equal to 5 in and 3 in each. Therefore, it is a parallelogram.

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x = 59,53445508

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Step-by-step explanation:

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svetlana [45]
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3 years ago
If x = a cosθ and y = b sinθ , find second derivative
Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

\dfrac{\mathrm d^2y}{\mathrm dx^2}

Compute the first derivative. By the chain rule,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}

We have

y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta

x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta

and so

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta

Now compute the second derivative. Notice that \frac{\mathrm dy}{\mathrm dx} is a function of \theta; so denote it by f(\theta). Then

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}

By the chain rule,

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}

We have

f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta

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\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta

4 0
3 years ago
1. Find the percent markup.
krek1111 [17]

\mathbb{ \ \underline{ \:  \:  ANSWER  : }}

1. Find the percent markup.

Markup ($) = percent markup (%) x original cost ($)

A computer store buys a laptop for $635 directly from the manufacturer (Dell). The markup is $570. What is the percent markup on the computer?

The percent markup is: <u> </u><u>90</u><u> </u>%

\mathbb{ \ \underline{ \:  \:  SOLUTION    : }}

FORMULA: Percent of Markup = Markup/Cost x 100

  • = $570/$635 x 100
  • = 0.90 x 100
  • = 90%

7 0
2 years ago
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