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Maurinko [17]
3 years ago
11

What is an equation of the line through point (2,1) and perpendicular to the line (-4,1) and (3, -2)?

Mathematics
1 answer:
MissTica3 years ago
8 0

Answer:

Gradient of the line with points (-4,1)&(3,-2) is

(-2-1)÷(3-(-4))=-3/7

Then the equation of that^ line is

Y-1=-3/7(x-(-4))--->y=-3/7x-5/7

Since this^ line is perpendicular to the line you are looking for.the gradient of the line you are looking for is -1÷(-3/7)=7/3

The equation of the line you are looking for is

Y-1=7/3(x-2)--->y=7/3x-11/3

[Cause the line you are looking for passes through the point (2,1) and you found the gradient of this line]

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Solve (2/3)-(1/x)=5/6
Keith_Richards [23]
\frac{2}{3} - \frac{1}{x} = \frac{5}{6}   Multiply both sides by 3
2 - \frac{1(3)}{x} = \frac{5(3)}{6}   Simplify the numerators
2 - \frac{3}{x} = \frac{15}{6}   Multiply both sides by 6
12 - \frac{3(6)}{x} = 15   Simplify the numerator
12 - \frac{18}{x} = 15   Multiply both sides by x
12x - 18 = 15x   Subtract 12x from both sides
       -18 = 3x    Divide both sides by 3
        -6 = x      Switch the sides to make it easier to read
         x = -6

Check your answer by plugging -6 in for the x in the original problem:

\frac{2}{3} - \frac{1}{x} = \frac{5}{6}   Plug in -6 for x
\frac{2}{3} - \frac{1}{-6} = \frac{5}{6}   Change \frac{2}{3} to \frac{4}{6} so all denominators are the same
\frac{4}{6} - \frac{1}{-6} = \frac{5}{6}   Change \frac{1}{-6} to  \frac{-1}{6} for easier work
\frac{4}{6} - \frac{-1}{6} = \frac{5}{6}   Subtract the numerators (4 - (-1) )
\frac{5}{6} = \frac{5}{6}

So, x = -6 .
5 0
3 years ago
(a) Plot the following function ona Karnaugh map.(Do not expand to minterm form before plotting.)
makvit [3.9K]

Answer:

a) the K-map is in the attachment

<em>f = Σ</em>m(0,1,2,3,6,10,14,15)

b) from the k-map, the minimum sum of products is

F = A'B' + CD' + ABC

c) the minimum product of sums is

F = (B' + C)(A' + C)(A+ B' +D')(A' + B + D')

Step-by-step explanation:

A Karnaugh map (K-map) is a pictorial framework used to limit the Boolean expressions without utilizing Boolean algebra theorems and equation controls.

a) the given function is <em>f</em>(A,B,C,D)=A‘B’+CD’+ABC +A’B’CD’+ABCD’

expanding the function as four variable terms

<em>f</em>(A,B,C,D)=A‘B’+CD’+ABC +A’B’CD’+ABCD’

= A'B'(C + C')(D + D')+(A + A')(B + B")CD' + ABC(D + D') + A'B'CD' + ABCD'

= A'B'CD + A'B'CD' + A'B'C'D' + ABCD' +AB'CD' + A'BCD' + A'B'CD' + ABCD +ABCD' + A'B'CD' + ABCD'

=A'B'CD + A'B'CD' + A'B'C'D + A'B'C'D' + ABCD' + AB'CD' + A'BCD' +ABCD

<em>f = Σ</em>m(0,1,2,3,6,10,14,15)

note: diagram is in the attachment

b) the minterms for the minimum sum of product are

<em>f = Σ</em>m(0,1,2,3,6,10,14,15)

simplifying the K-map(done in the attachment)

from the k-map, the minimum sum of products is

F = A'B' + CD' + ABC

c) the maxterms for the minimum product of sums are

<em>f </em>= ПM(4,5,7,8,9,11,12,13)

plot the K-map to find minimum product of sums(done in the attachment)

the minimum product of sums is

F = (B' + C)(A' + C)(A+ B' +D')(A' + B + D')

4 0
3 years ago
Find the measurement of either the leg or the hypotenuse: A = B = 21 C = 0<br> 35
Anarel [89]
I believe the answer is 21
5 0
3 years ago
Solve the system of equations by substitution
scoundrel [369]

Answer:

(x, y) = (1, 3)

Step-by-step explanation:

given the 2 equations

x + y = 4 → (1)

y = 3x → (2)

Substitute y = 3x into (1)

x + 3x = 4

4x = 4 ( divide both sides by 4 )

x = 1

Substitute x = 1 into (2) for corresponding value of y

y = 3 × 1 = 3

solution is (1, 3 )


7 0
3 years ago
How do I simplify the fraction 125/186
pogonyaev
125/186 cannot be simplified. It is already in simplest form.
6 0
4 years ago
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