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3 years ago

Give the equation of a line in slope-intercept form with the following criteria:

Mathematics

1 answer:

Vsevolod [243]3 years ago

3 0

Answer:

$y = - \frac{7}{6}x + 9$

Step-by-step explanation:

We need to find a straight line that is perpendicular to the line $y = \frac{6}{7} x + 4$ .

So, the slope of the given straight line is $\frac{6}{7}$ {Since the equation is in slope-intercept form}

Now, the slope of the required straight line will be [tex]- \frac{7}{6}[/tex]

{Since, the product of slopes of two straight line that are perpendicular to each other is -1, and $\frac{6}{7} \times (- \frac{7}{6}) = - 1$ }

Then the equation of the required straight line in slope-intercept form will be $y = - \frac{7}{6} x + c$ ............. (1) {Where c is any constant}

Now, point (12, -5) will satisfy the equation (1).

Hence, $-5 = - \frac{7}{6}(12) + c$

⇒ - 5 = - 14 + c

⇒ c = 9

Therefore, the complete equation of the required straight line is $y = - \frac{7}{6}x + 9$ (Answer)

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Solve for n<br> 4n + 12=36

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Answer:

n = 6

Step-by-step explanation:

4n + 12=36

Subtract 12 from each side

4n + 12-12=36-12

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3 years ago

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Well the answer is 14 cm, because all you have to do is divide the length of LN in half

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Consider the parabola r(t)equalsleft angle at squared plus 1 comma t right angle, for minusinfinityless thantless thaninfinity

kodGreya [7K]

Given:- $r(t)=< at^2+1,t>$ ; $-\infty < t< \infty$ , where a is any positive real number.

Consider the helix parabolic equation :

$r(t)=< at^2+1,t>$

now, take the derivatives we get;

$r{}'(t)=$

As, we know that two vectors are orthogonal if their dot product is zero.

Here, $r(t) and r{}'(t)$ are orthogonal i.e, $r\cdot r{}'=0$

Therefore, we have ,

$< at^2+1,t>\cdot < 2at,1>=0$

$< at^2+1,t>\cdot < 2at,1>=$

$=2a^2t^3+2at+t$

$2a^2t^3+2at+t=0$

take t common in above equation we get,

$t\cdot \left (2a^2t^2+2a+1\right )=0$

⇒ $t=0$ or $2a^2t^2+2a+1=0$

To find the solution for t;

take $2a^2t^2+2a+1=0$

The number $D = b^2 -4ac$ determined from the coefficients of the equation $ax^2 + bx + c = 0.$

The determinant $D=0-4(2a^2)(2a+1)$ $=-8a^2\cdot(2a+1)$

Since, for any positive value of a determinant is negative.

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David did not notice the multiplication sign between two three-digit numbers and wrote one six-digit number, which is equal to s

IRISSAK [1]

1) We can get that number hit and trial method.

It is said that "two three-digit numbers and wrote one six-digit number, which is equal to seven times their product".

Let us check number 143.

If we write 143 two times without any sign in between , we get 143143 ( a six digit number).

But if we multiply 143 × 143 , we get 20449.

7 times of 20449 equals 143143.

<h3>Therefore, the required three digit number is 143.</h3>

2) Let us assume required prime numbers are a, b and c.

Product of a, b and c= abc.

Sum of a, b and c = a+b+c.

"Their product is five times greater than their sum."

Therefore,

abc = 5(a+b+c) ----------------------equation (1)

Now, let us take first prime numbers 2 and second 5.

Plugging a=2 and b=5.

2×5×c = 5(2+5+c).

10c = 5(7+c)

10c = 35 +5c.

Subtracting 5c from both sides, we get

10c-5c = 35 +5c-5c.

5c = 35.

Dividing both sides by 5, we get

c=7.

Therefore, first "cool" triple is 2,5,7.

Let us check by taking a=2 and b=7.

Plugging a=2 and b=7 in equation (1), we get

2×7×c = 9(2+7+c).

c=9. But it's not a prime number.

Let us take a=2 and b=11, we get

2×11×c = 11(2+11+c).

c=13 (A prime)

If we take a=2 and b=17, we get

2×17×c = 17(2+17+c).

c=19 (A prime).

<h3>On the same way, if we continue the process, we can get many "cool" triples.</h3>

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