Asking again for your help! I really don't understand!

Answers

Answer 1

Answer:

Answer: y≤4x+3 and y≤-3x-5

Step-by-step explanation:

The slope of the first line is 4. The slope of (-1,-1) and (0,3) is 4. And the line passes through 3, so the y-intercept is 3.

The slope of the second line is 3. The slope of (-3,4) and (-1,-2) is -3. And the line passes through -5, so the y-intercept of the second line is -5

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20 points!!! please help

Solve each system of equations using substitution. Show all work.1) x + y = 1
2x - y = -2

2) 5x - 3y = -11
x - 2y = 2

3) x - y = 3
6x + 4y = 13

Answers

it would be so much easier with elimination but ok
1.
x+y=1
minus x
y=1-x
sub
2x-(x-1)=-2
2x-x+1=-2
x+1=-2
x=-3
sub
y=1-x
y=1-(-3)
y=1+3
y=4
(-3,4)

2.
x-2y=2
add 2y
x=2y+2
sub
5(2y+2)-3y=-11
10y+10-3y=-11
7y+10=-11
7y=-21
y=-3
sub
x=2y+2
x=2(-3)+2
x=-6+2
x=-4
(-3,-4)

3.
x-y=3
add y
x=y+3
sub
6(y+3)+4y=13
6y+18+4y=13
10y+18=13
10y=-5
y=-1/2
sub
x=y+3
x=-1/2+3
x=5/2

(5/2,-1/2)

1. (-3,4)
2. (-3,-4)
3.(5/2,-1/2)

The formula for the atrapezoid is Ah bi b2). Express bi in number of feet inarea of terms the of a trapezoid is 60 square feet, its height is 6 ft, and one base is 12 Find other base.

Answers

Area = 1/2h(b1 + b2)

60 = 1/2(6)(b1 +12)
60 = 3(b1 +12)
60 = 3b1 + 36
24 = 3b1
24/3 = b1

b1 = 8

The other base is 8 feet

Michael is 121212 years older than Brandon. Seventeen years ago, Michael was 444 times as old as Brandon. How old is Michael now?

Answers

Let Brandon's age be = b years

Let Michael's age be = m years

As Michael is 12 years older than Brandon, Michael's age, so

$m=b+12$

Seventeen years ago, Michael was 4 times as old as Brandon, so

$m-17=4(b-17)$

$m-17=4b-68$

$m=4b-51$

$m-4b=-51$

Substitute m=b+12

$b+12-4b=-51$

$-3b=-63$

$b=21$

Hence, Brandon's age is 21 years

As m=b+12

$m=21+12$

$m=33$

Hence, Michael's age is 33 years.

Use Synthetic Division to Factor the following polynomials completely by given factors or roots and Find all the zeros; When you reach quadratic equation, performance regular factoring or Quadratic Formula. $x^(5) -9x^(3) -x^(2) +9;(x-3),(x-1)$ , as factors and -3 as a root

Answers

Answer:

The factors are (x - (1 - i√3)/2) , (x - (1 + i√3)/2) , (x - 3) , (x + 3) , (x - 1)

The zeroes are 1 , 3 , -3 , (1 - i√3)/2 , (1 + i√3)/2

Step-by-step explanation:

$(x^(5)+0-9x^(3) -x^(2)+0+9)$ ÷ (x - 3) =

$(x^(4)-9x^(3)-x^(2)+0+9)$ ÷ (x - 3) =

$x^(4)+3x^(3)+(-x^(2)+0+9)$ ÷ (x - 3) =

$x^(4)+3x^(3)-x+(-3x+9)$ ÷ (x - 3) =

$(x^(4)+3x^(3)-x-3)$

$(x^(4)+3x^(3)+0-x-3)$ ÷ (x - 1) =

x³ + (4x³ + 0 - x - 3) ÷ (x - 1) =

x³ + 4x² + (4x² - x - 3) ÷ (x - 1) =

x³ + 4x² + 4x + (3x - 3) ÷ (x - 1) =

x³ + 4x² + 4x + 3

∵ -3 is a root ⇒ (x + 3) is a factor

(x³ + 4x² + 4x + 3) ÷ (x + 3) =

x² + (x² + 4x + 3) ÷ (x + 3) =

x² + x + (x + 3) ÷ (x + 3) =

x² + x + 1 ⇒ use the formula to find the factors of this quadratic

∵ a = 1 , b = 1 and c = 1

∴ $x=\frac{-1+\sqrt{(1)^(2)-4(1)(1)} }{2(1)}=(-1+√(-3))/(2)=(-1+i√(3))/(2)$

∴ $x=(-1-i√(3) )/(2)$

∴ The factors are (x - (1 - i√3)/2) , (x - (1 + i√3)/2) , (x - 3) , (x + 3) , (x - 1)

The zeroes:

x - 3 = 0 ⇒ x = 3

x + 3 = 0 ⇒ x = -3

x - 1 = 0 ⇒ x = 1

x - (1 - i√3)/2 = 0 ⇒ x = (1 - i√3)/2

x - (1 + i√3)/2 = 0 ⇒ x = (1 + i√3)/2

The zeroes are 1 , 3 , -3 , (1 - i√3)/2 , (1 + i√3)/2

Explain why the conversion ratios used between the SI and imperial systems are not nice, round numbers like those used to convert within either system.

Answers

Because within each system, units are defined in terms of others in the same system using nice round numbers, but no unit in either system is defined in terms of a unit in the other system. The one exception is the unit of time, and there, the relationship is a nice round number: 1 metric second = 1 imperial second.

A shoemaker bought 30 pairs of heels. How many individual heels did the shoemaker buy?a. 60
b. 15
c. 30
d. 120

Answers

30 pairs and a pair is 2 so 30*2 =60, so the answer is A

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