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3 years ago

Form the intersection for the following sets.

1 answer:

5 0

The intersection is where the sets overlap. picture a Venn diagram (two overlapping circles). The intersection is only the overlap part. So for this problem that would be C. {20}

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What’s the GCM of 16and 48 answer fast

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Write a quadratic function in standard form with a leading coefficient of 1 for the given set of zeros. 7 and 4

Viefleur [7K]

We are given : Zeros x=7 and x=4 and leading coefficent 1.

In order to find the quadratic function in standard form, we need to find the factors of quadratic function first and the multiply by given leading coefficent.

For the given zeros x=7 and x=4, we get the factors (x-7) and (x-4).

So, we need to multiply (x-7) and (x-4) by foil method.

We get

(x-7)(x-4) = x*x + x* -4 -7*x -7*-4

x^2 -4x -7x +28.

Combining like terms, we get

-4x-7x = -11x

x^2 -4x -7x +28 = x^2 -11x +28.

Now, we need to multiply x^2 -11x +28 quadratic by leading coefficent 1.

We get

1(x^2 -11x +28) = x^2 -11x +28.

Therefore, the required quadratic function in standard form is x^2 -11x +28.

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3 years ago

Please answer question three and four if you can :)<br> Show full working out ty;)

nikklg [1K]

Step-by-step explanation:

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

$D = ( \frac{2 + 5}{2}, \: \: \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \: \: \frac{ 1}{2} ) \\ \therefore D= (3.5, \: \: 0.5) \\ \& \: A=(-1,\:\:4)...(given) \\\\ now \: by \: distance \: formula \\ \\ Length \: of \: segment \: AD \\ = \sqrt{( - 1 - 3.5)^{2} + {(4 - 0.5)}^{2} } \\ = \sqrt{(4.5)^{2} + {(3.5)}^{2} } \\ = \sqrt{20.25 + 12.25 } \\ = \sqrt{32.5} \\ \red{ \boxed{\therefore Length \: of \: segment \: AD = 5.7 \: units}}$

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

$\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\$

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

$LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS$

Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

4 0

4 years ago