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3 years ago

Please help I don’t know the answer :(

Mathematics

2 answers:

gogolik [260]3 years ago

6 0

Answer:

When x=8, y=-6. When x=10, y=0

Step-by-step explanation:

$y=3x-30\\x^2+y^2=100$

You can plug the given y value into the second equation to get:

$x^2+(3x-30)^2=100\\x^2+9x^2-180x+900=100\\10x^2-180x+800=0\\10(x^2-18x+80)=0\\10(x-8)(x+10)=0\\x= 8, 10\\y= -6, 0$

Hope this helps!

GarryVolchara [31]3 years ago

4 0

-6,0 is the answer for this

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2 years ago

When X =10, the ??? Of the expression 4x is 40

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3 0

3 years ago

The diagram below shows a pennant in the shape of an isosceles triangle. The equal sides each measure 13. The altitude is x + 7

Kisachek [45]

Answer: The length of the base is 10.

Step-by-step explanation:

You can draw an isosceles triangle like the one shown attached.

An isosceles triangle has two equal sides. Then, you can divide it into two right triangles, where the base of each one of them will be:

$b'=\frac{2x}{2}=x$

The hypotenuse of each right triangle shown is 13 and the other legs are (x+7) and x, therefore you can apply the Pythagorean Theorem:

$13^2=x^2+(x+7)^2$

Simplify the expression:

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Use the quadratic formula, where:

$a=2\\b=14\\c=-120$

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

Then:

$x=\frac{-14\±\sqrt{14^2-4(2)(-120)}}{2+2}$

$x_1=5\\x_2=-12$

Choose the positive value, because lengths are not negative.

Substitute the value of x into the base 2x. Therefore, the answer is:

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7 0

3 years ago

All of these ODEs model a system with a spring, mass and dashpot.

Wewaii [24]

Answer:

− 3 y ' ' − 3 y ' + 3 y = 0 : over-damped

− 2 y ' ' − 4 y ' + 1 y = 0 : over-damped

1 y ' ' + 7 y ' + 5 y = 0: over-damped

Step-by-step explanation:

Using the characteristic equation you can express a differential equation of order n as an algebraic equation of degree n:

$a_ny^n+a_n_-_1y^{n-1}+...+a_1y'+a_oy=0$

This differential equation will have a characteristic equation of the form:

$a_nr^n+a_n_-_1r^{n-1}+...+a_1r+a_o=0$

Now, you can classify the solution for a differential equation using a simple method. In order to do it, you just need to use the discriminant.

If the discriminant is greater than zero, the solution is over-damped

If the discriminant is less than zero, the solution is under-damped

If the discriminant is equal to zero, the solution is critically damped

So, given the differential equation:

$-3y''-3y+3y=0$

Which has characteristic equation of the form:

$-3r^2-3r+3=0$

The quadratic polynomial of the form:

$ar^2+br+c=0$

Has discriminant:

$Disc=b^2-4ac$

In this case:

$a=-3\\b=-3\\c=3$

So:

$Disc=(-3)^2-4(-3)(3)=9-(-36)=45$

In this case:

$Disc=45>0$

Therefore the solution is over-damped.

Now, given the differential equation:

$-2y''-4y'+1y=0$

Which has characteristic equation of the form:

$-2r^2-4r+1=0$

The quadratic polynomial of the form:

$ar^2+br+c=0$

Has discriminant:

$Disc=b^2-4ac$

In this case:

$a=-2\\b=-4\\c=1$

So:

$Disc=(-4)^2-4(-2)(1)=16+8=24$

In this case:

$Disc=24>0$

Therefore the solution is over-damped.

Finally, given the differential equation:

$1y''+7y'+5y=0$

Which has characteristic equation of the form:

$1r^2+7r+5=0$

The quadratic polynomial of the form:

$ar^2+br+c=0$

Has discriminant:

$Disc=b^2-4ac$

In this case:

$a=1\\b=7\\c=5$

So:

$Disc=(7)^2-4(1)(5)=49-20=29$

In this case:

$Disc=29>0$

Therefore the solution is over-damped.

6 0

3 years ago