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3 years ago

A rectangular lunch tray is 33 cm long and 26 cm wide.

Mathematics

1 answer:

beks73 [17]3 years ago

3 0

Answer:

The area of the 858 square centimeters.

Step-by-step explanation:

You multiply 33 by 26 and that gives you the area. You can use the equation lw=a

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Find the sum of the first 30 multiples of 8.

Anna11 [10]

Step-by-step explanation:

sum of the arithmetic sequence

a1 = 8

a2 = a1 + 8 = 8+8 = 16

...

an = 8×n

the sum of the first n elements of such a series is

n/2 × (2×a1 + (n - 1)×d]

with d being the difference between 2 consecutive elements (= the added constant).

so,

s30 = 30/2 × (2×8 + 29×8) = 15×(31×8) = 3720

3 0

3 years ago

A calculator may be used for this question.

Lilit [14]

14 x 3/4 = 42/4 = 10.5 lbs
1 lb = 450 g...so 10.5 lbs = (10.5 x 450) = 4725 g.....1 g = 0.001 kg....so 4725 grams = (4725 x .001) = 4.725 kg

so ur gonna need a medium sized turkey.....4.725 kg...at 5.99 per kg =
(4.725 x 5.99) = $ 28.30 <==

7 0

4 years ago

Question 1

vivado [14]

Answer:

33, 792

Step-by-step explanation:

first convert the 8 miles to feet because our answers are in feet.

1 mile= 5280 feet

therefore 8 miles= 5280×8

= 42240

Dale and son covered 4/5 of the 42240 feet

= 4/5×42240

= 33,782 feet

4 0

2 years ago

What is 44 in simplified radical form? A.11/4 B.11/2 C.4/11 D.2/11

Aleksandr-060686 [28]

C.4/11 is the answer to your question

6 0

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