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Marizza181 [45]

2 years ago

11

Look at the image below for the question !!!!!!

1 answer:

Semenov [28]2 years ago

4 0

Answer:

136 cm2

Step-by-step explanation:

6*2=12 12*2=<u>24</u>

6*7=42 42*2=<u>84</u>

7*2=14 14*2=<u>28</u>

Add.

24+84+28=136

I hope this helps!

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Melissa reserves another rectangular patch in her vegetable garden for growing bell peppers. Since the space that she can reserv

Otrada [13]

Answer:

Quadratic, 16, 18, 6

Step-by-step explanation:

Did this assignment myself

7 0

3 years ago

Which expressions are equivalent to (d^1/8) ^5

Tanya [424]

Answer:

Formula used : (a^m)^n = a^(m×n)

$→{( {d}^{ \frac{1}{8} }) }^{5} = {d}^{( \frac{1}{8}\times 5) } = \boxed{{d}^{ \frac{5}{8} } }✓ \\$

<u>d^⅝</u> is the right answer.

8 0

2 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

A factory produce bicycles at a rate of 110+0.5t^2-0.9t bicycles per week (t in weeks)

klasskru [66]

Answer:

221

Step-by-step explanation:

Day 8 is the first day of the second week.

Day 21 is the last day of week 3.

We need to know the n umber of bicycles made from t = 1 to t = 3

The function is b(t) = 110 + 0.5t^2 - 0.9t, where t is in weeks.

We need to integrate the function with the limits of 1 to 3.

$\int_{1}^{3} (110 + 0.5t^2 - 0.9t) dt$

$\int_{1}^{3} (110 + \dfrac{t^2}{2} - \dfrac{9t}{10}) dt$

$= 110t + \dfrac{t^3}{6} - \dfrac{9t^2}{20} \Biggr|_{1}^{3}$

$= 330.45 - 109.72$

$= 220.7333$

Answer: 221

3 0

3 years ago

Please help me! Please!

user100 [1]

Answer:

C) -3 is the answer

6 0

2 years ago