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3 years ago

7

Which could be the area of a face of the rectangular prism?

1 answer:

Dmitry_Shevchenko [17]3 years ago

5 0

Answer:

The answer is

B. 6m²

Step-by-step explanation:

N/B: the drawing was not provided, so kindly find attached the drawing for your reference.

We know that the area of a rectangle = length *breath

From the diagram the rectangular surface has a length l=3m

Breath b=2m

Area = 3*2= 6m²

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Lisa and her daughter weigh 185 pounds all together. The ratio of lisa’s weight to her daughter weight is 4:1 . How much does li

zzz [600]

Since the ratio of mother to daughter is 4:1, the weight of the mother and daughter together is five times the weight of the daughter.

In fact, if the daughter weights $d$ pound, the mother weights $4d$ pounds, and the total weight is $d+4d = 5d$ .

We know that this total is 185 pounds, so we have

$5d = 185 \iff d = \dfrac{185}{5} = 37$

So, the daughter weights 37 pounds.

6 0

3 years ago

What is the fractional equivalent of 0.2323?

nekit [7.7K]

2323/1000=4646/2000
<span> is the fractional equivalent of 0.2323</span>

3 0

3 years ago

Read 2 more answers

Likurg_2 [28]

Answer:

$(10+2p)/p=((20+0.80p)/p)+1$

Step-by-step explanation:

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3 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

3 years ago

Please help- I cannot get this wrong

mr_godi [17]

Answer:

Top left table

Step-by-step explanation:

It increases at a constant rate of 2 in the y for every 2 in the x so its slope is 1.

7 0

3 years ago

Read 2 more answers