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Alexxandr [17]
3 years ago
11

there are 100 pennies in a dollar what fraction of a dollar is 61 pennies write a fraction decimal and word form

Mathematics
1 answer:
ycow [4]3 years ago
5 0
\bf \begin{array}{ccllll}
dollar&pennies\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
1&100\\
x&61
\end{array}\implies \cfrac{1}{x}=\cfrac{100}{61}

solve for "x", that's how much
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Solve the given inequality and graph the solution on a number line -x/2 + 3/2 + < 5/2 .
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The ground temperature at sea level is 60 degrees F. For every 100-foot increase in elevation, the temperature rises 1/10 of one
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<h3>Answer: B. 62 degrees fahrenheit</h3>

====================================================

Explanation:

x = elevation in feet

y = temperature in fahrenheit

The temperature goes up 1/10 = 0.1 degrees for every 100-foot increase of elevation. So the slope is 0.1/100 = 0.001, which tells us how fast the temperature is increasing. In other words, the temperature goes up 0.001 degrees each time the elevation goes up by 1 foot.

The ground temperature is 60 degrees, which is our starting temperature. It's the value of y when x = 0. Therefore, 60 is the y intercept.

We have a slope of m = 0.001 and a y intercept of b = 60. The equation y = mx+b becomes y = 0.1x+60

Now plug in x = 2000 to find the temperature at this elevation

y = 0.001x+60

y = 0.001*2000+60

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Use the distributive property to write an equivalent expression: 5(3x + 2)
LenKa [72]

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Suppose that f: R --&gt; R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su
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<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

f(x+y)=f(x)+f(y)------(1)  ∀  x,y ∈ R

Now, let us assume f(1)=k

Also,

  • f(0)=0

(  Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0  )

Also,

  • f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1)         ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

  • Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}

Also,

  • when x∈ Q

i.e.  x=\dfrac{p}{q}

Then,

f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq belonging to Q such that:

\to x )

Now, we know that: Q'=R

This means that:

Э α ∈ R

such that Э sequence a_n such that:

a_n\ belongs\ to\ Q

and

a_n\to \alpha

f(a_n)=ka_n

( since a_n belongs to Q )

Let f is continuous at x=α

This means that:

f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha

This means that:

f(\alpha)=k\alpha

                       This means that:

                    f(x)=kx for every x∈ R

4 0
3 years ago
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