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3 years ago

We bought five 3 3/4 pound bags of bird seed. What is the total weight of the birdseed they bought

Mathematics

1 answer:

Firdavs [7]3 years ago

3 0

It would be 18.75 because if you put it into a mixed number then multiply by 5 you get 18.75

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What is X squared +3x-70

Aleksandr-060686 [28]

Answer:

x = 7, x = -10

Step-by-step explanation:

$x^2+3x-70 = 0$

Use the quadratic formula.

$x = \frac{-3 + \sqrt{3^2-4(1)(-70)} }{2(1)} \\x = \frac{-3 - \sqrt{3^2-4(1)(-70)} }{2(1)} \\$

Solve.

x = 7, x = -10

You can also factor if you want - that is a faster method.

8 0

3 years ago

WILL AWARD BRAINLIEST AND 30 PTS IF THERE IS A GOOD EXPLANATION. ANSWER IS NOT 243. question: In how many ways can we put five i

Bumek [7]

Answer:

243 ways

Step-by-step explanation:

Fruits = 5

Bowls = 3

Remember that, There are three bowls for five fruits. So the possible outcomes are:

=> $3^5 = 3*3*3*3*3 = 243$ ways

8 0

3 years ago

I need help on this question plz

soldi70 [24.7K]

Cylinders are 3D figures, as they are solids with volume and take up space.

2D figures are flat figures that you can draw, like a square or circle.

1D figures are actually just a line, a segment, or a point, etc.

8 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

3 years ago

Last question I need help with an answer not a link

Lapatulllka [165]

Answer:

it's toooo complicated

6 0

2 years ago