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4 years ago

HELP

1 answer:

Ber [7]4 years ago

4 0

Paralell has same slope
the slope for ax+by=c is -a/b
that means that a line paralell to 2x+3y=3 will have 2x+3y=something
find that something

given
(3,-4)
x=3 and y=-4 is a soltuion
2x+3y=something
2(3)+3(-4)=something
6-12=something
-6=something

2x+3y=-6

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Solve x² + 11x +18=0 by factoring.

kirill [66]

-(11+sqrt(121-4*18))/2 = -(11 +sqrt(49))/2 = -9

Or

-(11-sqrt(121-4*18))/2 = -(11-sqrt(49))/2 = -2

D is the answer

5 0

3 years ago

An airplane flies due north from ikeja airport for 500km.It then flies on a bearing of 060 from a further distance of 300km befo

Nata [24]

Answer:

482 km

63.94 degrees

Step-by-step explanation:

to solve this question we will use the cosine rule. For starters, draw your diagram. From point A, up north is 500km and 060 from there, another 300. If you join the point from the road junction back to the starting point, yoou have a triangle.

Cosine rule states that

C = $\sqrt{A^{2} + B^{2} -2AB cos(c) }$

where both A and B are the given distances, 500 and 300 respectively, C is the 3rd distance we're looking for and c is the given angle, 060

solving now, we have

C = $\sqrt{500^{2} + 300^{2} -2 * 500 * 300 cos(60) }$

C = $\sqrt{250000 + 90000 - [215000 cos(60) }]$

C = $\sqrt{340000 - [215000 * 0.5 }]$

C = $\sqrt{340000 - [107500 }]$

C = $\sqrt{232500}$

C = 482 km

The bearing can be gotten by using the Sine Rule.

$\frac{sina}{A}$ = $\frac{sinc}{C}$

sina/500 = sin60/482

482 sina = 500 sin60

sina = $\frac{500 sin60}{482}$

sina = 0.8983

a = sin^-1(0.8983)

a = 63.94 degrees

6 0

3 years ago

Please help me with this math question

Leona [35]

Answer:

X + Y - W - Z = 180

Step-by-step explanation:

In parallel lines If you use the rules of:

* Alternate angles are equal

* Linear Pair Angles supplement to 180

And use the picture below for calculating the result: finally we get

Z = Y - (180 - (X - W)) which becomes the answer above if you simplify the parentheses.

5 0

3 years ago

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

zheka24 [161]

Answer:

$f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

Step-by-step explanation:

Let:

$M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y$

This is and exact equation, because:

$\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}$

So, define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)$

The solution will be given by:

$f(x,y)=C_1$

Where C1 is an arbitrary constant

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y):

$f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)$

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}$

Substitute into $\frac{\partial f(x,y)}{\partial y} =N(x,y)$

$2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y$

Integrate $\frac{dg(y)}{dy}$ with respect to y:

$g(y)=\int\ {8y} \, dy =4y^2$

Substitute g(y) into f(x,y):

$f(x,y)=2x^2+4y^2+2xy$

The solution is f(x,y)=C1

$f(x,y)=2x^2+4y^2+2xy=C_1$

Solving y using quadratic formula:

$y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

4 0

4 years ago

What is 90% of 50? Someone please help. I have to write this for my question to get a good answer.

irina1246 [14]

45÷50=90%
if that is what you are looking for

8 0

3 years ago

Read 2 more answers