Graph by making a table. y= -8x

Answers

Answer 1

Answer: x y
-5 40
-4 32
-3 24
-2 16
-1 8
0 0
1 -8
2 -16
3 -24
4 -32
5 -40

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How do you factorise 14x -21xy

Answers

Simplifying 14x + -21xy

Factor out the Greatest Common Factor (GCF), '7x'. 7x(2 + -3y)

Final result:
7x(2 + -3y)

that is the answer on how to do it

You need to find the common number/letter in order to factorise and it should be like this:
x(14-21y)
Hope it helped...

Solve for x : 2x^2+4x-16=0

Answers

$2x^2+4x-16=0\ \ \ /:2\n\nx^2+2x-8=0\n\n\underbrace{x^2+2x\cdot1+1^2}_((*))-1^2-8=0\n\n(x+1)^2-1-8=0\n\n(x+1)^2-9=0\n\n(x+1)^2=9\iff x+1=-3\ or\ x+1=3\n\nx=-3-1\ or\ x=3-1\n\nx=-4\ or\ x=2\n\n\n(*)\ (a+b)^2=a^2+2ab+b^2$

$2x^2+4x-16=0\ \ \ /:2\n\nx^2+2x-8=0\n\na=1;\ b=2;\ c=-8\n\n\Delta=b^2-4ac\ if\ \Delta > 0\ then\ x_1=(-b-\sqrt\Delta)/(2a)\ and\ x_2=(-b+\sqrt\Delta)/(2a)\n\n\Delta=2^2-4\cdot1\cdot(-8)=4+32=36;\ \sqrt\Delta=√(36)=6\n\nx_1=(-2-6)/(2\cdot1)=(-8)/(2)=-4;\ x_2=(-2+6)/(2\cdot1)=(4)/(2)=2$

in order to solve it, we need find the zero of the polynomial.

we find the zero of the polynomial by splitting the middle term method

2x2 -+ 4x - 16

= 2x2 + 8x - 4x -16

= 2x( x + 4)- 4(x + 4)
= (2x-)(x+4)

we find the zeroes of the factors

experimentally we find two values, 2 and -4.

Thus, values of x are 2 and -4

Which of the two functions below has the smallest minimum y-value?f(x) = x4 - 2
g(x) = 3x3 + 2

Answers

Answer:

out of the two functions g(x) attains the minimum y-value.

Step-by-step explanation:

We are given the function $f(x)=x^4-2$ and $g(x)=3x^3+2$ .

now we have two check out of the two functions above which has the smallest minimum y-value.

Clearly from the graph attached to the solution of f(x) and g(x), we could see that for f(x) the minimum value of y is -2 while for g(x) the value of y goes on decreasing as x goes to minus infinity.

Hence g(x) attains the minimum value for y.

Quadrilateral EFGH has coordinates E(a, 2a), F(3a, a), G(2a, 0), and H(0, 0). Find the midpoint of HG.A (2a, 0)
B (a, 2a)
C (a, a)
D (a, 0)

Answers

Midpoint of HG = ((0 + 2a)/2, (0 + 0)/2) = (2a/2, 0/2) = (a, 0)
option D is the correct answer.

Coordinates are: H ( 0, 0 ) and G ( 2 a, 0 ).
Coordinates of the midpoint are:
x = ( x 1 + x 2 ) / 2
y = ( y 1 + y 2 ) / 2
The midpoint of HG:
x = ( 0 + 2 a ) / 2 = 2 a / 2 = a
y = ( 0 + 0 ) / 2 = 0 / 2 = 0
Answer:
D ) ( a, 0 )

NEED answer soon. just the 4 letters thankss

Answers

Answer:

a = 25m^2

b = 5m

d = 35.73 m^2

c = 7.94m

Step-by-step explanation:

First, remember that the area of a square of side length L is:

A = L^2

And for a triangle rectangle with catheti a and b, and hypotenuse H, we have the relation:

H^2 = a^2 + b^2 (Phytagorean's theorem)

Ok, let's look at the left image, we have a green triangle rectangle.

The bottom cathetus has a length equal to the side length of a square with area of 16m^2

Then the side length of that square (and the cathetus) is:

L^2 = 16m^2

L = √(16m^2) = 4m

The left cathetus has a length equal to the side length of a square of area = 9m^2

Then the side length of that cathetus is:

K^2 = 9m^2

K = √(9m^) = 3m

Then the catheti of the green triangle rectangle are:

4m and 3m

Then the hypotenuse of this triangle (b) is:

b^2 = (4m)^2 + (3m)^2

b^2 = 16m^2 + 9m^2 = 25m^2

b = √(25m^2) = 5m

And b is the side length of the red square, then the area of that square is:

a = b^2 = 25m^2

Now let's go to the other image.

Here we have an hypotenuse of side length H, such that:

H^2 = 144m^2

And we have a cathetus (the one adjacent to the green triangle) of side length L such that:

L^2 = 81m^2

The other cathetus will have a sidelength c, such that:

c^2 = area of the purple square

By the Pythagorean's theorem we have:

144m^2 = 81m^2 + c^2

144m^2 - 81m^2 = c^2

63m^2 = c^2

(√63m^2) = c = 7.94m

And the area of a triangle rectangle is equal to the product between the catheti divided by two.

We know that one cathetus is equal to c = 7.94m

And the other on is equal to the square root of 81m^2

√(81m^2) = 9m

then the area of the triangle is:

d = (7.94m)*(9m)/2 = 35.73 m^2

Please help me with this one
10x+5y=72
15x+4y=80

Answers

I hope this helps you

Problem: Solve 10x+5y=72;15x+4y=80 Steps: I will try to solve your system of equations. 10x+5y=72;15x+4y=80 Step: Solve10x+5y=72for x: 10x+5y+−5y=72+−5y(Add -5y to both sides) 10x=−5y+72 10x 10 = −5y+72 10 (Divide both sides by 10) x= −1/2 y+36 5 Step: Substitute −1 2 y+ 36 5 forxin15x+4y=80: 15x+4y=80 15( −1 2 y+ 36 5 )+4y=80 −7 2 y+108=80(Simplify both sides of the equation) −7 2 y+108+−108=80+−108(Add -108 to both sides) −7 2 y=−28 −7 2 y −7 2 = −28 −7 2 (Divide both sides by (-7)/2) y=8 Step: Substitute8foryinx= −1 2 y+ 36 5 : x= −1 2 y+ 36 5 x= −1 2 (8)+ 36 5 x= 16 5 (Simplify both sides of the equation) Answer: x=16/5 and y=8

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