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3 years ago

Math help What are 3 question about the equation do you need to ask before solving?

Mathematics

1 answer:

tatuchka [14]3 years ago

4 0

What strategies can i use to solve this equation

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Find the 31st term of the sequence 12,16,20

Natali [406]

Answer:

136

Step-by-step explanation

multiply 4 by 31 and add 20

6 0

3 years ago

What is the least common denominator used to add the fractions 1/18 and 5/12

Ne4ueva [31]

Answer:

Step-by-step explanation:

Im right

7 0

3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

2 years ago

Find the determinant of the following matrix. A =<br>-2 6<br>3 5

Blababa [14]

Answer:

$|A| = - 28$

Step-by-step explanation:

Given

$A = \left[\begin{array}{cc}-2&6\\3&5\end{array}\right]$

Required

Determine the determinant

For a two by two matrix, A such that:

$A = \left[\begin{array}{cc}a&b\\c&d\end{array}\right]$

The determinant |A| is:

$|A| = a * d - b * c$

So, in

$A = \left[\begin{array}{cc}-2&6\\3&5\end{array}\right]$

The determinant is:

$|A| = -2 * 5 - 6 * 3$

$|A| = -10 - 18$

$|A| = - 28$

6 0

2 years ago

3/5+(-7/5)= can someone help me with this please

Tresset [83]

Answer:

$-\frac{4}{5}$

Step-by-step explanation:

Because both of the fractions have a common denominator they can be subracted without any other work.

4 0

2 years ago