The lines 2x + ky = 3 and kx + 8y = 7 intersect for all values of k except 4 and -4 the answer is R - {4, -4}.
It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.
If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.
It is given that:
Two linear equations in two variables are:
2x + ky = 3 and
kx + 8y = 7
As we know the condition for the lines to intersect or have unique solutions:
2/k ≠ k/8
After cross multiplication:
k² ≠ 16
k ≠ ±4
The values of k can be anything except 4 and - 4
Thus, the lines 2x + ky = 3 and kx + 8y = 7 intersect for all values of k except 4 and -4 the answer is R - {4, -4}.
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Answer:
ky=3-2x
y=(3-2x)/k
y=(7-kx)/8
(3-2x)/k=(7-kx)/8
8(3-2x)=(7-kx)k
24-16x=7k-k*kx
24=7k-kkx+16x
24+7k=kkx+16x
24+7k=x(k^2+16)
The fraction as an improper fraction is 38/9
From the question, we have the following parameters that can be used in our computation:
Fraction = 4 2/9
When converted to an improper fraction, we have
Numerator = 9 * 4 + 2
Evaluate
Numerator = 38
The denominator remains the same
So, we have
Fraction = 38/9
Hence, the fraction is 38/9
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Answer:
Step-by-step explanation:
For a quadratic equation, the vertex form is given by : , where (h, k) is the vertex.
The given quadratic equation:
Subtract 9 from both sides
compare this to , and add both sides
b= 20
[(b/2)²=20/2=10]
So, the vertex form :