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3 years ago

Home skool or reg skool (whatever that is considered nowadays)

Mathematics

2 answers:

lutik1710 [3]3 years ago

5 0

Answer:

home school

Step-by-step explanation:

svp [43]3 years ago

4 0

Answer:home school or online school.

Step-by-step explanation:

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Madison represented the sentence

Angelina_Jolie [31]

Answer:

The mistake is that it can be 5. They forget to include the equals in the less than or equals

Step-by-step explanation:

Start at the inside and work out

quotient of a number and -2

n/-2

difference of -4 and the quotient of a number and -2

-4 - n/-2

product of 3 and the difference of -4 and the quotient of a number and -2

3* (-4 - n/-2)

is at most 5

3* (-4 - n/-2)≤5

The mistake is that it can be 5. They forget to include the equals in the less than or equals

3 0

3 years ago

Read 2 more answers

I need to finish this question ASAP Someone please help!

Helga [31]

Answer:

Step-by-step explanation:

g(r) = 0 ⇒ $r^{2} -2r-17=0$ ⇒r= $1+3\sqrt{2}$ và r= $1-3\sqrt{2}$

4 0

3 years ago

22: A 2.0 kg object moves along the x-axis under the influence of an unknown force. Its potential energy due to the force is U(x

aleksandr82 [10.1K]

Answer: a) -13.25 N, b) 1.78 m/sec

Step-by-step explanation:

Since we have given that

$U(x)=ax^4+bx^2$

where

$a=2.5\ J/m^4\\\\b=12\ J/m^2$

As we know that

$F=\dfrac{-dU}{dx}=-(4ax^3+2bx)$

Here, x= 0.5 m

So, it becomes,

$F=-(4\times 2.5\times (0.5)^3+2\times 12\times 0.5)=-13.25\ N$

Now, we need to find the velocity when it crosses the origin,

$F=mg\\\\F=m\dfrac{dv}{dt}\\\\F=mv\dfrac{dv}{dx}\\\\-(4ax^3+2bx)dx=mvdv\\\\\int\limits^{0.5}_0 {-(4ax^3+2bx)} \, dx =\int\limits {mv} \, dv\\\\ \dfrac{-4ax^4}{4}+\dfrac{2bx^2}{2}|_0^{0.5}=\dfrac{mv^2}{2}\\\\ax^4+bx^2|_0^{0.5}=\dfrac{mv^2}{2}\\\\2.5(0.5)^4+12(0.5)^2=\dfrac{2v^2}{2}\\\\0.15625+3=v^2\\\\3.15625=v^2\\\\\sqrt{3.15625}=v\\\\v=1.78$

Hence, a) -13.25 N, b) 1.78 m/sec

7 0

4 years ago

C=1/21.22.23+1/22.23.24+................+1/200.201.202 . = là dấu nhân

Aneli [31]

It looks like you have to find the value of the sum,

$C = \displaystyle \frac1{21\times22\times23} + \frac1{22\times23\times24} + \cdots + \frac1{200\times201\times202}$

so that the n-th term in the sum is

$\dfrac1{(21+(n-1))\times(21+n)\times(21+(n+1))} = \dfrac1{(n+20)(n+21)(n+22)}$

for 1 ≤ n ≤ 180.

We can then write the sum as

$\displaystyle C = \sum_{n=1}^{180} \frac1{(n+20)(n+21)(n+22)}$

Break up the summand into partial fractions:

$\dfrac1{(n+20)(n+21)(n+22)} = \dfrac a{n+20} + \dfrac b{n+21} + \dfrac c{n+22}$

Combine the fractions into one with a common denominator and set the numerators equal to one another:

$1 = a(n+21)(n+22) + b(n+20)(n+22) + c(n+20)(n+21)$

Expand the right side and collect terms with the same power of n :

$1 = a(n^2+43n+462)+b(n^2+42n+440) + c(n^2+41n + 420) \\\\ 1 = (a+b+c)n^2 + (43a+42b+41c)n + 462a+440b+420c$

Then

a + b + c = 0

43a + 42b + 41c = 0

462a + 440b + 420c = 1

==> a = 1/2, b = -1, c = 1/2

Now our sum is

$\displaystyle C = \sum_{n=1}^{180} \left(\frac1{2(n+20)}-\frac1{n+21}+\frac1{2(n+22)}\right)$

which is a telescoping sum. If we write out the first and last few terms, we have

C = 1/(2×21) - 1/22 + 1/(2×23)

… … + 1/(2×22) - 1/23 + 1/(2×24)

… … + 1/(2×23) - 1/24 + 1/(2×25)

… … + 1/(2×24) - 1/25 + 1/(2×26)

… … + … - … + …

… … + 1/(2×198) - 1/199 + 1/(2×200)

… … + 1/(2×199) - 1/200 + 1/(2×201)

… … + 1/(2×200) - 1/201 + 1/(2×202)

Notice the diagonal pattern of underlined and bolded terms that add up to zero (e.g. 1/(2×23) - 1/23 + 1/(2×23) = 1/23 - 1/23 = 0). So, like a telescope, the sum collapses down to a simple sum of just six terms,

C = 1/(2×21) - 1/22 + 1/(2×22) + 1/(2×201) - 1/201 + 1/(2×202)

which we simplify further to

C = 1/42 - 1/44 - 1/402 + 1/404

C = 1,115/1,042,118 ≈ 0.00106994

4 0

3 years ago

What is the degree of the following monomial? 5fg^3h^5

12345 [234]

Answer:

The degree is one for f, g, and h.

Step-by-step explanation:

8 0

3 years ago

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