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Tju [1.3M]
2 years ago
6

What percent of 17 is 3?

Mathematics
1 answer:
Vikki [24]2 years ago
4 0

Answer: 3 is 17.6471% of 17

Step-by-step explanation:

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PLEASE HELP ME I BEG YOU!!
ivann1987 [24]

1) false

2)true

3)-1.346

<span>4)<span><span>x=−5</span><span>x=-5</span></span></span>

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<span>6)x =~ 3.91 </span>

<span>7) x = 162.69 </span>

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7)

6 0
3 years ago
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What is the expression for "the sum of 8 times a number and two"?
LiRa [457]

Answer:

8 N + 2

Step-by-step explanation:

==> 8 N + 2

-----------_________-------------

8 0
2 years ago
What are factors of 42 that subtract to -8
MrRa [10]

You might have to do the quadratic formula

8 0
3 years ago
Is bse~ tes? if so, identify the similarity the similarity postulate or theorem that applies
hammer [34]
The triangles are congruent by the SAS congruence theorem. By extension, we can also say they are similar using a closely related theorem. Any time there are congruent triangles, they are automatically similar.

This is why the answer is choice C) Similar - SAS

Note how
SB = ET ... which forms the first "S" in "SAS"
angle ESB = angle SET ... which forms the "A" in "SAS"
SE = SE ... which forms the second "S" in "SAS"
this is why SAS works
8 0
3 years ago
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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
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