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3 years ago

Need help with science

Mathematics

1 answer:

kogti [31]3 years ago

3 0

Answer:

where is the question?

Step-by-step explanation:

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Solve for X y and z brainliest Round to nearest tenth

kykrilka [37]

Answer:

x=1

Z=35.4965

assuming you want y to be the height, y=5.916

Step-by-step explanation:

x/height=height/35

35x=y^2

using Pythagorean theorem

y^2+x^2=6^2

y^2+35^2=z^2

substituting for y^2

35x+x^2=6^2

x^2+35x-36=0

(x+36)(x-1)=0

x=1

y^2+x^2=6^2

substitute

y^2+1=36

y^2=35

y=5.916

y^2+35^2=z^2

substitute

35+1225=z^2

z^2=1260

z=35.4965

4 0

3 years ago

A 41 gram sample of a substance that's used to detect explosives has a k-value of 0.1392. Find the substance's half life, in day

-BARSIC- [3]

Answer:

The substance half-life is of 4.98 days.

Step-by-step explanation:

Equation for an amount of a decaying substance:

The equation for the amount of a substance that decay exponentially has the following format:

$A(t) = A(0)e^{-kt}$

In which k is the decay rate, as a decimal.

k-value of 0.1392.

This means that:

$A(t) = A(0)e^{-0.1392t}$

Find the substance's half life, in days.

This is t for which $A(t) = 0.5A(0)$ . So

$A(t) = A(0)e^{-0.1392t}$

$0.5A(0) = A(0)e^{-0.1392t}$

$e^{-0.1392t} = 0.5$

$\ln{e^{-0.1392t}} = \ln{0.5}$

$-0.1392t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.1392}$

$t = 4.98$

The substance half-life is of 4.98 days.

5 0

3 years ago

Arte-miy333 [17]

Answer:

if you have a bad credit score it can increase your debt ratio

Step-by-step explanation:

3 0

4 years ago

Which one of these is the correct unit of measurement in finding volume?

kirill [66]

Answer:

feet cubic

Step-by-step explanation:

8 0

3 years ago

Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks

Iteru [2.4K]

Let

$P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}$

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

$P(1):\ 1 = \dfrac{1\cdot 2}{2}=1$

So, the base case is ok. Now, we need to assume $P(n)$ and prove $P(n+1)$ .

$P(n+1)$ states that

$P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}$

Since we're assuming $P(n)$ , we can substitute the sum of the first n terms with their expression:

$\underbrace{1+2+\ldots+n}_{P(n)}+n+1 = \dfrac{n(n+1)}{2}+n+1=\dfrac{n(n+1)+2n+2}{2}=\dfrac{n^2+3n+2}{2}$

Which terminates the proof, since we showed that

$P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}$

as required

4 0

3 years ago