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Sati [7]
3 years ago
5

Write a equation of the lines passes through the given points (-2,7) (1,1) the equation is?

Mathematics
1 answer:
vredina [299]3 years ago
7 0

The point-slope form:

y-y_1=m(x-x_1)

m - slope

(x₁, y₁) - point

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (-2, 7) and (1, 1). Substitute:

m=\dfrac{1-7}{1-(-2)}=\dfrac{-6}{3}=-2

y-7=-2(x-(-2))\\\\\boxed{y-7=-2(x+2)}   <em>point-slope form</em>

\boxed{y=-2x+3}     <em> slope-intercept form</em>

\boxed{2x+y=3}       <em>standard form</em>

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Answer:  angle 1 = 110 degrees

This is because angle 1 and the red 110 degree angle are corresponding angles.

They correspond to one another due to the fact they are both in the southwest corner of each 4-corner angle configuration (which is the result of crossing two lines)

Corresponding angles are congruent only when we have parallel lines like this.

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3 years ago
Please help. If this isn't done by today i'm grounded :(
DedPeter [7]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
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If a cup of coffee has temperature 97°C in a room where the ambient air temperature is 25°C, then, according to Newton's Law o
FinnZ [79.3K]

Answer:

The  average temperature  is T_{a} = 81.95^oC

Step-by-step explanation:

From the question we are told that

    The temperature of the coffee after time t is   T(t) =  25 + 72 e^{[-\frac{t}{45} ]}

Now the average temperature during the first 22 minutes i.e fro 0 \to  22minutes is mathematically evaluated as

              T_{a} =  \frac{1}{22-0}  \int\limits^{22}_{0} {25 +72 e^{[-\frac{t}{45} ]}} \, dx

               T_{a} = \frac{1}{22} [25 t  +  72 [\frac{e^{[-\frac{t}{45} ]}}{-\frac{1}{45} } ] ] \left| 22} \atop {0}} \right.

             T_{a} = \frac{1}{22} [25 t  - 3240e^{[-\frac{t}{45} ]} ] \left | 45} \atop {{0}} \right.

              T_{a} = \frac{1}{22} [25 (22)  - 3240e^{[-\frac{22}{45} ]}   - (- 3240e^{0} )]

            T_{a} = \frac{1}{22} [550  - 1987.12  +  3240]

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without carrying a tank bottoms has the size of 3 to 6 m. Find out how many cans of paint of 2 kg are required to paint the insi
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3 years ago
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Determine whether the sequences converge.
Alik [6]
a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}

Notice that

\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}

So as n\to\infty you have a_n\to0. Clearly a_n must converge.

The second sequence requires a bit more work.

\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then a_n will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When n=2, you have

a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1

Assume a_k\ge a_{k-1}, i.e. that a_k=\sqrt{2a_{k-1}}\ge a_{k-1}. Then for n=k+1, you have

a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k

which suggests that for all n, you have a_n\ge a_{n-1}, so the sequence is increasing monotonically.

Next, based on the fact that both a_1=\sqrt2=2^{1/2} and a_2=2^{3/4}, a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}
a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}

and so on. We're getting an inkling that the explicit closed form for the sequence may be a_n=2^{(2^n-1)/2^n}, but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, a_1=2^{1/2}. Let's assume this is the case for n=k, i.e. that a_k. Now for n=k+1, we have

a_{k+1}=\sqrt{2a_k}

and so by induction, it follows that a_n for all n\ge1.

Therefore the second sequence must also converge (to 2).
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3 years ago
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