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2 years ago

Make x the subject of the formula 6x+a=5

Mathematics

1 answer:

Tatiana [17]2 years ago

7 0

You would -a to get 6x=5-a then divide by 6 which you would get your final answer which is x=5-a/6

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(x^3- 2x^2 - 1) /( x - 1)

lesya [120]

Answer:

Quotient= x^2 + x + 1

Remainder= 0

Step-by-step explanation:

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3 years ago

A mean average of 70 or better on five exams is needed for a final grade of C. Jeffery Matthews first four exam grades are 75, 6

erastova [34]

Answer:38

Step-by-step explanation:

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3 years ago

10tu^3 + 4v - 3v<br><br> must combine like terms solve or rewrite expression

pshichka [43]

10tu³ + v is the simplified form.

<h3>What is algebraic expression?</h3>

A number, a variable, or a mix of numbers, variables, and operation symbols make up an expression. Two expressions joined by an equal sign form an equation. Word illustration: The product of 8 and 3. Word illustration: The product of 8 and 3 is 11

to learn more about algebraic expression refer to:

brainly.com/question/4344214

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1 year 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

3 years ago

Question in the photo below

vichka [17]

Answer:

Step-by-step explanation:

Well the slope is y = -3/2x and B is the only option with that slope.

5 0

3 years ago

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