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3 years ago

What is the area of a rectangle with a length of 7.5 feet and a width of 8 feet?

Mathematics

2 answers:

Lerok [7]3 years ago

6 0

The area of a rectangle is length*width so the area of this rectangle is 7.5 * 8 or 60ft²

Pepsi [2]3 years ago

4 0

Answer:

d 60 ft 2

Step-by-step explanation:

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2. What is the center of dilation and the scale factor of the dilation of ABCD to EFGH?

Allushta [10]

Answer:

The center of dilation of ABCD to EFGH is (0, 0), and the scale factor is 4/3

6 0

3 years ago

The recursive formula of an recursive sequence is described below a1=20, an-1+6

Mandarinka [93]

Answer:

The answer is 17.80

Step-by-step explanation:

Just took the test hope this helps:)

3 0

4 years ago

In a bag of 100 marbles comprised of many colors, 25 are blue. suppose marbles are selected at random.

satela [25.4K]

Solution: The number of ways we can arrange 3 blue marbles if a set of 5 marbles is selected is:

$\binom{5}{3}=\frac{5!}{(5-3)!3!}$

$=\frac{120}{2 \times 6}$

$=\frac{120}{12}=10$

Therefore, there are 10 ways we could arrange 3 blue marbles.

3 0

3 years ago

J&J Manufacturing issued a bond with a $1,000 par value. The bond has a coupon rate of 7% and makes payments semiannually. I

Andreyy89

Answer:

The correct option is d. $ 785

Step-by-step explanation:

Since,

$\text{Bond price}=\frac{C}{YTM}(1-\frac{1}{(1+\frac{YTM}{2})^{2M}})+\frac{FV}{(1+\frac{YTM}{2})^{2M}}$

Where,

C = Annual coupon payment,

FV = Face value,

M = Maturity in years,

YTM = yield to maturity,

Here,

FV = $ 1,000,

C = 7% of 1000 = $\frac{7\times 1000}{100}$ = 70,

M = 20 years,

YTM = 9.4% = 0.094,

By substituting the values,

$\text{Bond price}=\frac{70}{0.094}(1-\frac{1}{(1+\frac{0.094}{2})^{40}})+\frac{1000}{(1+\frac{0.094}{2})^{40}}$

= $ 785.3454 ( Using calculator )

≈ $ 785

Hence, OPTION d. is correct.

8 0

3 years ago

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

$\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

$\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}$

7 0

3 years ago