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4 years ago

Can someone help me with all this...

Mathematics

1 answer:

professor190 [17]4 years ago

4 0

Answer:

Option D -Parallelogram and rhombus is the correct answer.

Step-by-step explanation:

Option A -Square.

In square all sides are equal and all angles are equal and each angle is equal to 90°

Option B - Parallelogram

In Parallelogram opposites sides are equal and parallel. there is no right angles.

Option C -Rectangle

In rectangle opposites sides are equal and parallel. All angles are equal to 90°

Option D -Parallelogram and rhombus

In Parallelogram opposites sides are equal and parallel. there is no right angles.

In rhombus all sides are equal and parallel. there is no right angles.

Therefore correct answer is option D

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Step-by-step explanation:

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3 years ago

Enter an inequality that represents the phrase the sum of 1 and y is greater than or equal to 5.

Delicious77 [7]

Sum means we are adding. 1 + y has to be greater than or equal to 5, therefore, $1+y \geq 5$

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3 years ago

Roger's Cafe has usual daily earnings of $800. Today, the cafe earned 120% of the usual daily earnings. How much did they earn t

riadik2000 [5.3K]

Answer:

960

Step-by-step explanation:

Over here, I'm going to do it in two ways, and either way works, one is just easier to do out:

1st way:

You can represent the percent 120 as a decimal which is 1.20, then you do 800*1.2.

2nd way:

You can use benchmark percents. You find 10 percent of 800 which is 80 and multiply 80 by 2 to get 160, so 20% is 160 dollars. Then you add 160 to 800 (800+160) and get 960.

Please tell me if you have any questions.

8 0

3 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$