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2 years ago

What is an equivalent expression for 40+36

1 answer:

7 0

An equiavalent expression for 40+36 can be:

46+30
50+26
etc.

As long as the expression equals 76, it is correct

hope this helps

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plz relpy asap can you tell me these shapes and if they are quadrilateral ,a kite,trapezoid, parallelogram or, rhombus, rectangu

OleMash [197]

Not sure about 1 and 8, I think there trapezoids but not sure.

2 is a square

3 is a rectangle

4 and 10 are a rhombus

5 and 11 are parallelograms

6 is a kite

7 is a square

can't really see 9

can't really see 12

Hope that helped.

4 0

2 years ago

A recipe calls for 2 1/5 cups of chopped tomatoes and 3 2/5 cups of diced turnips how many more cups of turnips did the recipe c

Zarrin [17]

The answer would be 1 1/5 more turnips.

6 0

2 years ago

Read 2 more answers

The beads in 7 bracelets if 4 bracelets have 96 bead

Naya [18.7K]

96/4=24 beads in 1 bracelet
beads in 7 bracelets = 24*7=168

3 0

2 years ago

One-third a number (x) added to one-half the number equals nine less than the number

Anna11 [10]

The answer to your question is:

1/3x + 1/2x = x - 9

8 0

2 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