All categories

3 years ago

Two lines intersect in _____.

Mathematics

1 answer:

yarga [219]3 years ago

4 0

The correct answer to this question is C. One Point.

You might be interested in

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