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3 years ago

Which rectangle could be created from a parallelogram with a base of 6 and a height of 2?

Mathematics

1 answer:

nalin [4]3 years ago

5 0

Answer:

A rectangle with a length of 6 and width of 2.

The corners added up make a rectangle so it is the same as a rectangle. You can just multiply 6 x 2 = 12 to find the area. If you have any more questions you are welcome to ask

Step-by-step explanation:

You might be interested in

Amanda uses 1/3 of a cup of milk each time she makes a batch of pancakes. How manybbatches can she make if she only has 11/22 of

kow [346]

Answer:

<u>Amanda can make 1.5 or 3/2 of batches of pancakes with 11/22 of a cup of milk.</u>

Step-by-step explanation:

1. Let's review the information given to us for answering the question correctly:

Amount of milk Amanda use for a batch of pancakes = 1/3 cup

2. How many batches can she make if she only has 11/22 of a cup of milk left?

Let's simplify 11/22, dividing by 11 the numerator and the denominator,

11/22 = 1/2

Now, using the Rule of Three Simple, we can calculate the answer to this question, this way:

Number of batches = (1/2 * 1)/1/3

Number of batches = 1/2 / 1/3

Number of batches = 1/2 * 3/1 (Taking the reciprocal)

Number of batches = 3/2 = 1.5

<u>Amanda can make 1.5 or 3/2 of batches of pancakes with 11/22 of a cup of milk.</u>

4 0

3 years ago

If x is directly proportional to to y and x= 200 and y=3 find x when y=5

Amiraneli [1.4K]

Answer:

333.33

Step-by-step explanation:

x is directly proportional to y

k=y/x

k=3/200

now,

when y =5

y=kx

5=(3/200)x

1000/3=x

333.33 = x

3 0

2 years ago

Simplify completely (2x^2+16x+30)/(5x^2+13x-6)

spin [16.1K]

Answer:

2(x+5) / 5x-2

Step-by-step explanation:

(2x^2+16x+30)/(5x^2+13x-6)

6 0

3 years ago

Find the value of x when 2(x - 3) + 1 = 3x + 4

GREYUIT [131]

2x-6+1=3x+4
2x-5=3x+4
2x-3x=4+5
-x=9

8 0

3 years ago

Read 2 more answers

F(x) = (128/127)(1/2)x, x = 1,2,3,...7. determine the requested values: round your answers to three decimal places (e.g. 98.765)

Marrrta [24]

A.
$\mathbb P(X\le 1)=\mathbb P(X=1)=\dfrac{128}{127}\left(\dfrac12\right)^1=\dfrac{64}{127}$

b.
$\mathbb P(X>1)=1-\mathbb P(X\le1)=1-\dfrac{64}{127}=\dfrac{63}{127}$

c.
$\mathbb E(X)=\displaystyle\sum_{x=1}^7 x\,f_X(x)=\frac{64}{127}\sum_{x=1}^7 x\left(\frac12\right)^{x-1}$

Suppose $f(y)=\displaystyle\sum_{x=0}^7 y^x$ . Then $f'(y)=\displaystyle\sum_{x=1}^7 xy^{x-1}$ . So if we can find a closed form for $f(y)$ , in terms of $y$ , we can find $\mathbb E(X)$ by evaluating the derivative of $f(y)$ at $y=\dfrac12$ .

$f(y)=\displaystyle\sum_{x=0}^7 y^x=y^0+y^1+y^2+\cdots+y^6+y^7$
$y\,f(y)=y^1+y^2+y^3+\cdots+y^7+y^8$
$f(y)-y\,f(y)=y^0-y^8$
$(1-y)f(y)=1-y^8$
$f(y)=\dfrac{1-y^8}{1-y}$
$\implies f'(y)=\dfrac{7y^8-8y^7+1}{(1-y)^2}$
$\implies\mathbb E(X)=\dfrac{64}{127}f'\left(\dfrac12\right)=\dfrac{64}{127}\times\dfrac{247}{64}=\dfrac{247}{127}$

d.
$\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$

We find $\mathbb E(X^2)$ in a similar manner as in (c).

$\mathbb E(X^2)=\displaystyle\sum_{x=1}^7 x^2\,f_X(x)=\frac{32}{127}\sum_{x=1}^7x^2\left(\frac12\right)^{x-2}$

Now,

$f(y)=\displaystyle\sum_{x=0}^7y^x$
$\implies f'(y)=\displaystyle\sum_{x=1}^7xy^{x-1}$
$\implies f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}$

We know that

$f''(y)=-\dfrac{42y^8-96y^7+56y^6-2}{(1-y)^3}$
$\implies f''\left(\dfrac12\right)=\dfrac{219}{16}$

We also have

$f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}$
$f''(y)=\displaystyle\sum_{x=2}^7x^2y^{x-2}-\sum_{x=2}^7xy^{x-2}$
$f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=2}^7x^2y^x-\sum_{x=2}^7xy^x\right)$
$f''(y)=\displaystyle\frac1{y^2}\left(\bigg(\sum_{x=1}^7x^2y^x-y\bigg)-\bigg(\sum_{x=1}^7xy^x-y\bigg)\right)$
$f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=1}^7x^2y^x-\sum_{x=1}^7xy^x\right)$

so that when $y=\dfrac12$ , we get

$\dfrac{219}{16}=4\left(\dfrac{127}{128}\mathbb E(X^2)-\dfrac{127}{128}\mathbb E(X)\right)\implies\mathbb E(X^2)=\dfrac{685}{127}$

Then

$\mathbb V(X)=\dfrac{685}{127}-\left(\dfrac{247}{127}\right)^2=\dfrac{25,986}{16,129}$

6 0

3 years ago