All categories

2 years ago

Ab+ d= c+ e solve for a

Mathematics

1 answer:

blondinia [14]2 years ago

8 0

Answer:

a=e

Step-by-step explanation:

a+abcdef b=asd d+da

You might be interested in

Multiple Choice

Zigmanuir [339]

Answer:

the slope is POSITIVE (answer A)

Step-by-step explanation:

The line goes through: (-1, -2) and ( 2, 1)

Then the slope can be calculated as:

slope = (y2 - y1) / (x2 - x1)

in our case:

slope = (1 - -2) / (2 - -1) = (1 + 2) / (2 + 1) = 3 / 3 = 1

Therefore, the slope is POSITIVE (answer A)

8 0

2 years ago

Read 2 more answers

Carlos drunk 2740 milliliters after football practice how much liters does he drink

Anni [7]

2.74 liters since 1000mL is a liter.

7 0

3 years ago

Steve paid $62 for groceries last week. That was $6.75 more then he spent this week. How much did Steve spend this week?

yarga [219]

the answer is to 62-6.75 is 55.25

6 0

3 years ago

Read 2 more answers

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

Write an equation about having 6 puzzles, getting 5 more and having a lot now.

sattari [20]

6+5=11
or
6+5=?
It shows we originally have 6 and now have 5 more having a total of 11

8 0

3 years ago