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3 years ago

Three components of the term

Mathematics

1 answer:

jek_recluse [69]3 years ago

7 0

Answer: “5” is an exponent

Step-by-step explanation:

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3 5/12 + 1 3/8<br><br> I am to lazy i am needed for answer cause my eyes fell off

docker41 [41]

Answer:

4 19/24

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7 0

3 years ago

Find an expression which represents the difference when (2x − 6) is subtracted

Sloan [31]

Answer:

-2x + 8

Step-by-step explanation:

(2x - 6) is subtracted from (7 - 5).

First term that into an equation.

Because our first term is being subtracted from our second one, our equation is written as:

(7 - 5) - (2x - 6).

Simplify the first term by subtracting:

7 - 5 = 2

Now simplify the second term by multiplying 2x - 6 by -1 to get them out of the parenthesis. We do this because there is a negative sign in front of it.

-1 (2x - 6)

-2x + 6

Now our equation is:

2 - 2x + 6

Combine like terms:

2 + 6 = 8

So -2x + 8

(2x - 6) subtracted from (7 - 5) is -2x+8

4 0

1 year ago

Pls answer as soon as possible what is 5.3 rounded?

Nadusha1986 [10]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

If the Math Olympiad Club consists of 13 students, how many different teams of 5 students can be formed for competitions?

rjkz [21]

Answer:

2 teams

Step-by-step explanation:

13 divided by 5 is 2.6, but you cannot split a person in half, so it would be two teams with a remainder of three people.

4 0

3 years ago

Find the inverse Laplace transform f(t) of the function F(s). Write uc for the Heaviside function that turns on at c, not uc(t).

zzz [600]

Answer:

F(t) $=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$

Step-by-step explanation:

We have given $F(S)=\frac{7e^{-7s}}{s^2-49}$

Now $F(S)=e^{-7s}G(s)$

Here $G(S)=\frac{7}{S^2-49}$

Now first find the Laplace inverse of G(S)

Using partial fraction

$\frac{7}{(s+7)(s-7)}=\frac{A}{(S+7)}+\frac{B}{S-7}$

$7=A(S-7)+B(S+7)$

On comparing the coefficient

$A=\frac{1}{2}$ and $B=\frac{-1}{2}$

On putting the value of A and B

$G(S)=\frac{-1}{2(S+7)}+\frac{1}{2(S+7)}$

Taking inverse Laplace

$G(t)=\frac{-1}{2}e^{7t}+\frac{1}{2}e^{-7t}$

Now in G(s) there is onether term $e^{-7s}$

So F(t) $=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$

6 0

3 years ago

Three components of the term​

Three components of the term