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3 years ago

Find midpoint of 1,-3 and -1,-1

1 answer:

6 0

Find answer by using formula below and see correct answer with working in the picture below.
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Your Numbers are; 25, 75, 50, 1, 9, 3 and your Target is 386 How do i get to 386 using only these numbers once and I can use Sub

GuDViN [60]

Answer:

You must order the operations in the way that it unfolds in the explanation;:

Step-by-step explanation:

(25 x 9) +75 + 50+ 9 x ( 3 + 1)

When performing the operations we have the following results:

(25 x 9) =225 +75= 300+50 = 350+9 = 359 x (3+1 )

Now we must solve the following: 359 x (4) = 386

8 0

3 years ago

Pls solve I need it solved rn show work urgent PLS*****

omeli [17]

Answer:

16. y=-5x-20; 17. y=6x+18, or factored y=6(x+3)

Step-by-step explanation:

Your slope is -5

y=-5x+b, solve for b by pluging the point (-3,-5) in for x and y

-5=-5(-3)+b, solve for b

-20=b, now rewrite the equation

y=-5x-20

Your slope is 6

y=-5x+b, solve for b by pluging the point (-3,-0) in for x and y

0=6(-3)+b, solve for b

18=b, now rewrite the equation

y=6x+18, or factored y=6(x+3)

5 0

3 years ago

What is <br> 2x+y=-4<br> step by step please and i will give you 44 points for this!!!

Anni [7]

Answer: 0=0

Step-by-step explanation:

First you want to solve for x

2x+y=−4

Step 1: Add -y to both sides.

2x + y + −y =−4+ −y

2x=−y−4

Step 2: Divide both sides by 2.

2x/2= -y-4/2

x=-1/2y-2

Then you want to plug in your answer for x into the equation so:

2(-1/2y-2)+y=-4

Step 1: Simplify both sides of the equation.

2(-1/2y-2)+y=-4

(2)( −1/ 2 y)+(2)(−2)+y(Distribute)

−y+−4+y=−4

(−y+y)+(−4)=−4(Combine Like Terms)

−4=−4

Step 2: Add 4 to both sides.

−4+4=−4+4

0=0.

Sorry if this wasn't the answer your looking for. If you need more help I suggest using Ma.th.Pa.pa calculator but with out the periods. Hope you have a good day :)

5 0

3 years ago

Which points on the curve of x^2 - xy - y^2 = 5 have vertical tangent lines?

algol [13]

We need to differentiate this with respect to x to see if we can find an expression for the derivative of y at various points. That will be the slope of the tangent to the curve. Then we want to see where that derivative might be infinite -- i.e., where the tangent is vertical.

It's not written as a function, but it can still be differentiated using the chain rule:

x2 + xy + y2 = 3

(2x) + (x dy/dx + y dx/dx) + (2y dy/dx) = 0

(I used parentheses to show the differentiation of each term in the original equation.)

2x + x dy/dx + y + 2y dy/dx = 0

2x + y = -x dy/dx - 2y dy/dx

2x + y = dy/dx (-x -2y)

-(2x + y)/(x + 2y) = dy/dx

We have the derivative of y, but it's defined partly in terms of y itself. That's OK. Let's go on...

So where would the slope be infinite? That would happen when x + 2y = 0, or y = -x/2

Let's plug that in for y in the original equation to find points where that's the case.

x2 + xy + y2 = 3

x2 + x(-x/2) + (-x/2)2 = 3

x2 - x2/2 + x2/4 = 3

3x2 / 4 = 3

x2 = 4

x = ±2

So we have two x values where the tangent might be vertical. Let's plug them into the equation and see what the y values are. First x = 2...

x2 + xy + y2 = 3

4 + 2y + y2 = 3

y2 + 2y + 1 = 0

(y + 1)2 = 0

y = -1

So at the point (2, -1) the tangent is vertical.

Now try x = -2...

x2 + xy + y2 = 3

4 - 2y + y2 = 3

y2 - 2y + 1 =0

(y - 1)2 = 0

y = 1

So at the point (-2, 1) the tangent is vertical.

8 0

3 years ago

Fill in the missing statement and reason in the proof of the Alternate Exterior Angles Theorem.

ludmilkaskok [199]

Answer:

∠EGB and ∠AGF are congruent;

Transitive Property of Equality

Step-by-step explanation:

The vertical angles theorem is about the angles that are opposite to each other. These angles are formed when two lines cross each other

Vertical angles are congruent i.e their measures are equal

Look at the attached graph

Given:

AB // CD

E , G , H , F are col-linear

The line which contains points E , G , H , F is a transversal for the parallel lines AB and CD

∵ AB and EF intersected at point G

∴ m∠EGB = m∠AGF ⇒ by the vertical angles theorem

When lines AB and EF intersected at G they formed opposite angles like angles EGB and AGF, then angles EGB and AGF are congruent ( vertical angle theorem)

∵ AB // CD and EF is a transversal

∴ m∠AGF = m∠CHF (corresponding angles theorem)

Transitive Property of Equality: if one angle is equal to two other angles, then the two other angles are equal

Therefore, If a = b and b = c, then a = c

∵ m∠EGB = m∠AGF

∵ m∠AGF = m∠CHF

∴ m∠EGB = m∠CHF (transitive property theorem)

∠EGB and ∠AGF are congruent; Transitive Property of Equality

4 0

3 years ago

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