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quester [9]
3 years ago
15

Two lines intersect in _____.

Mathematics
1 answer:
yarga [219]3 years ago
4 0
The correct answer to this question is C. One Point.
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Write an equation of the line that passes through the given point and has the given slope.
Minchanka [31]

Answer:

y = -x - 4

Step-by-step explanation:

the equation of a linear graph is y = mx + b

you know the m = -1

so now the equation is y = -x + b

from here, what you could do to find b is...

plug (-1, -3) into the equation

1) x = -1, y = -3

2) -3 = -(-1)  + b

3) -3 = 1 + b

4) -4 = b

or you can look at the y-intercept which is where the line intersects with the y-axis

it seems to be (0, -4)

so "b" would be equal to -4

this makes the final equation y = -x + (-4) or y = -x - 4

4 0
3 years ago
Which set of integers is included in (-1,3]?
notka56 [123]
(1)
Cuz "[]" is in included
"()" isn't.
3 0
3 years ago
If Doris paid $24.30 for 8.1 pounds of Swiss cheese, what was the price of 1
lesantik [10]
It should be 3, I’m not sure
3 0
3 years ago
Prove that 1³+2³+....n³=n²(n+1)²/4. principle of mathematics induction​
alex41 [277]
<h3>The Simplified Question:-</h3>

\sf 1^3+2^3\dots n^3=\dfrac{n^2(n+1)^2}{2}

\\ \sf\longmapsto 1^3+2^3+\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2

<h3>Solution:-</h3>

Let

\\ \sf\longmapsto P(n)=1^3+2^3\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2

For n=1

\\ \sf\longmapsto P(1)=\left(\dfrac{1(1+1)}{2}\right)^2

\\ \sf\longmapsto P(1)=\left(\dfrac{1(2)}{2}\right)^2

\\ \sf\longmapsto P(1)=\left(\dfrac{2}{2}\right)^2

\\ \sf\longmapsto P(1)=(1)^2

\\ \bf\longmapsto P(1)=1=1^3

Let k be any positive integer.

\\ \sf\longmapsto P(k)= 1^3+2^3\dots k^3=\left(\dfrac{k(k+1)}{2}\right)^2

We have to prove that p(k+1) is true.

consider

\sf 1^3+2^3\dots k^3+(k+1)^3

\\ \sf\longmapsto \left(\dfrac{k(k+1)}{2}\right)^2+(k+1)^3

\\ \sf\longmapsto \dfrac{k^2(k+1)^2}{4}+(k+1)^3

\\ \sf\longmapsto \dfrac{k^2(k+1)^2+4(k+1)^3}{4}

\\ \sf\longmapsto \dfrac{k+1)^2\left\{k^2+4k+4\right\}}{4}

\\ \sf\longmapsto \dfrac{(k+1)^2(k+2)^2}{4}

\\ \sf\longmapsto \dfrac{(k+1)^2(k+1+1)^2}{4}

\\ \sf\longmapsto \left(\dfrac{(k+1)(k+1+1)}{2}\right)^2

\\ \sf\longmapsto (1^3+2^3+3^3\dots k^3)+(k+1)^3

Thus P(k+1) is true whenever P(k) is true.

Hence by the Principal of mathematical induction statement P(n) is true for \bf n\epsilon N.

Note:-

We can solve without simplifying the Question .I did it for clear steps and understanding .

<h3>Learn More:-</h3>

brainly.com/question/13253046?

brainly.com/question/13347635?

6 0
3 years ago
Read 2 more answers
Complete the pattern 5, 10, 30, 60, 180
saveliy_v [14]
5, 10, 30, 60, 180, 360, 1,080

5*2=10
10*3=30
30*2=60
60*3=180
180*2=360
360*3=1,080

6 0
3 years ago
Read 2 more answers
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