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3 years ago

Help with this please

Mathematics

1 answer:

Alik [6]3 years ago

7 0

Answer:

Yes, the given parallelogram is a rectangle.

Step-by-step explanation:

The vertices of parallelogram are J(-5,0), K(1,4), L(3,1) and M(-3,-3).

The slope formula is

$m=\frac{y_2-y_1}{x_2-x_1}$

$JK=\frac{4-0}{1-(-5)}=\frac{4}{6}=\frac{2}{3}$

$KL=\frac{1-4}{3-1}=\frac{-3}{2}$

$LM=\frac{-3-1}{-3-3}=\frac{-4}{-6}=\frac{2}{3}$

$JM=\frac{-3-0}{-3-(-5)}=\frac{-3}{2}$

The slopes of opposites sides are same it means they are parallel to each other.

The product of slopes of two consecutive sides is

$\frac{2}{3}\times \frac{-3}{2}=-1$

Since the product of slopes of two consecutive sides is -1, therefore the consecutive sides are perpendicular to each other.

Yes, the given parallelogram is a rectangle.

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Write an equation with the two points: (5,-2) and (3,4)

eimsori [14]

Answer:

$y=-3x-17$

Step-by-step explanation:

To find out the m(slope) of the equation use the formula:

$\frac{y_2-y_1}{x_2-x_1}$ .

Substitute the values:

$\frac{4--2}{3-5}$

Combine the values:

$\frac{6}{-2}$

Find the value of the fraction:

$-3$

M is -3.

The equation so far is $y=-3x+b$

To find the y intercept, substitute one pair of coordinates into the equation and solve for b:

$y=-3x+b\\-2=3*5+b\\-2=15+b\\-17=b$ .

We see that b =-17 so the full equation would be:

$y=-3x-17$ .

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2 years ago

Write the equation of a circle centered at the

Fudgin [204]

Answer:

$(x+4)^2+(y+6)^2=100$

Step-by-step explanation:

Given a circle centred at the point P(-4,-6) and passing through the point

R(2,2).

To find its equation, we follow these steps.

Step 1: Determine its radius, r using the distance formula

For point P(-4,-6) and R(2,2)

$\text{Distance Formula=}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\text{Radius=}\sqrt{(2-(-4))^2+(2-(-6))^2} \\=\sqrt{(2+4))^2+(2+6)^2}\\=\sqrt{6^2+8^2}\\=\sqrt{100}\\Radius=10$

Step 2: Determine the equation

The general form of the equation of a circle passing through point (h,k) with a radius of r is given as: $(x-h)^2+(y-k)^2=r^2$

Centre,(h,k)=P(-4,-6)

r=10

Therefore, the equation of the circle is:

$(x-(-4))^2+(y-(-6))^2=10^2\\\\(x+4)^2+(y+6)^2=100$

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3 years ago

Please explain with working!!! Find the set of values of x that satisfy the inequality 9x^2-15x<6

Oksi-84 [34.3K]

Answer:

$-1/3$

Step-by-step explanation:

When solving a quadratic inequality, first solve it normally like you would for a normal quadratic equation. We have:

$9x^2-15x$

Ignore the less than sign and replace it with an equal sign and solve the quadratic for its zeros:

$9x^2-15x=6$

Subtract 6 from both sides:

$9x^2-15x-6=0$

Divide everything by 3:

$3x^2-5x-2=0$

Factor. Find two numbers that equal (3)(-2)=-6 that add up to -5.

-6 and 1 works. Thus:

$3x^2-6x+x-2=0\\3x(x-2)+1(x-2)=0\\(3x+1)(x-2)=0$

Find the x using the Zero Product Property:

$3x+1=0 \text{ or }x-2=0\\x=-1/3\text{ or }x=2$

Now, we need to replace the equal signs with symbols again. To do so, we need to test which symbol to place. Let's do the first zero first.

So, the first zero is:

$x=-1/3$

Assume that the correct symbol is >. Thus,

$x>-1/3$

Now, pick any number that is greater than -1/3. I'll pick 0 since it's the easiest. Now, plug 0 back into the very original inequality. If it works, then the sign is correct, if it doesn't, then simply use the opposite one. Therefore:

$9x^2-15x$