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3 years ago

9

Decide whether ▱ABCD with vertices A(−2,3), B(2,3), C(2,−1), and D(−2,−1) is a rectangle, a rhombus, or a square. Select all nam

es that apply.
rectangle

rhombus

square

1 answer:

ozzi3 years ago

3 0

Answer:

D and B

Step-by-step explanation:

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Bella needs to read three chapters of her book. If she reads 3/5 of a chapter each night, how many nights will it take her to re

konstantin123 [22]

It will take her 5 nights to read three chapters.

Take $\frac{3}{5}$ and add up until you get to 3.

1st night: 3/5

2nd night: 1 1/5

3rd night: 1 4/5

4th night: 2 2/5

5th night: 3.

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3 years ago

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Write the system of equations, then solve. SHOW ALL WORK!! please HELP! QUICK

uysha [10]

Answer:

3 30 page printers, 10 70 page printers

Step-by-step explanation:

x + y = 13

40x+70y = 820

Solving for y in first equation gives: y = 13 - x

Plug into second equation: 40x + 70(13 - x) = 820

40x + 910 - 70x = 820

-30 x = -90

x = 3

Plug into original equation: 3 + y = 13

y must = 10.

So, the company owns 3 thirty page presses, and 10 70 page presses,

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2 years ago

What digit is in the ten thousands place of the number 43,310.60 975

Nadya [2.5K]

4 because you would say "forty three thousand..." three would be the one thousands place, and 4 would be the ten thousands place.

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4 years ago

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Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

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3 years ago

What is the least common denominator of the fractions 2/3, 1/4, 1/8 and ?

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Answer:

24

Step-by-step explanation:

That's the first number that all your denominators have in common.

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3 years ago