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djverab [1.8K]
3 years ago
9

Use substitution method & show your work.y = x^2 + 3x - 14x + 1 = y

Mathematics
1 answer:
irina1246 [14]3 years ago
7 0
4x + 1 = x^2 + 3x - 1
4x = x^2 + 3x - 2
0 = x^2 -x - 2
0 = x^2 -x + 1 - 3
0 = (x - 1)^2 - 3
3 = (x - 1)^2
sqrt 3=(x-1) and (-(x-1))
x = 1+sqrt 3 and 1-sqrt 3
sqrt means square root
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X^2+20x+11=-8 pls help me I need it now pls help
Rudiy27

Answer:

x = -19   or   x = -1

Step-by-step explanation:

x^2 + 20x + 11 = -8

We need the quadratic equation in the form

ax^2 + bx + c = 0,

so we add 8 to both sides.

x^2 + 20x + 19 = 0

Now we have an equation of the form x^2 + ax + b = 0.

We need to factor the quadratic equation. To factor it, we need two numbers, p and q, whose product is b and whose sum is a. Then the factoring is (x + p)(x + q).

In our case, b = 19 and a = 20. We need two numbers that multiply to 19 and add to 20. The numbers are 19 and 1. Our p and q are 19 and 1.

(x + 19)(x + 1) = 0

Since a product of two factors equals zero, either one factor equals zero, or the other factor equals zero. We set each factor equal to zero and solve both equations for x.

x + 19 = 0   or   x + 1 = 0

x = -19   or   x = -1

6 0
4 years ago
Please help me with both questions!!!!​​
Naya [18.7K]

Answer:

Step-by-step explanation:

16.

points are (0,-3),(2,1),(4,-3)

eq. of line through (0,-3) and (2,1) is

y+3=\frac{1+3}{2-0}(x-0)

or y+3=2x

2x-y=3

as the line is dotted so either < or >.

consider 2x-y>3

put x=0,y=0

0>3

which is not possible .

(0,0) does not satisfy the inequality

Hence shaded region is the required region.

now eq of line through (2,1)and (4,-3) is

y-1=\frac{-3-1}{4-2}(x-2)

y-1=-2(x-2)

y-1=-2x+4

2x+y=5

it is also a  dotted line so either < or >

consider 2x+y<5

put x=0,y=0

0<5

which is true.

so (0,0) satisfy this inequality.

so the two inequalities are

2x-y>3

and 2x+y<5

17.

consider the points (0,4),(2,3),(4,4)

eq. of line through (0,4) and (2,3) is

y-4=\frac{3-4}{2-0}(x-0)

y-4=-1/2(x)

2y-8=-x

or x+2y=8

as the line is solid

so either≤ or ≥

consider x+2y≥8

put x=0,y=0

0≥8

which is impossible.

(0,0) does not satisfy the graph.

which is true as graph lies above the line.

again eq. of line through (2,3) and (4,4) is

y-3=\frac{4-3}{4-2}(x-2)

y-3=1/2(x-2)

2y-6=x-2

x-2y=-4

consider x-2y≤-4

put x=0, y=0

0≤-4

which is impossible.

so (0,0) does not satisfy the graph.

so inequality is true as shaded portion is above and left of the line.

so two inequalities are

x+2y≥ 8

and x-2y≤-4

5 0
4 years ago
What words don’t describe? Super easy question!!!❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️
Mazyrski [523]

Answer:

1 - b

2 - d

3 - d

4 0
4 years ago
You need to reverse the symbol to solve which inequality?
Marina86 [1]

When you multiply/divide by a negative number while solving the inequality, you reverse the symbol.

A.)  4(5 - y) ≥ 80       To solve this, isolate/get the variable "y" by itself in the inequality. Divide 4 on both sides

\frac{4(5-y)}{4} \geq \frac{80}{4}

5 - y ≥ 20       Subtract 5 on both sides

5 - 5 - y ≥ 20 - 5

-y ≥ 15     Divide -1 on both sides to get "y" by itself

\frac{-y}{-1} \geq \frac{15}{-1}       Because you are dividing by a negative number, you need to reverse the sign

y ≤ -15

B.) 2(y - 3) ≥ 8      Isolate "y", divide 2 on both sides

y - 3 ≥ 4      Add 3 on both sides to get "y" by itself

y ≥ 7

C.) 3y - 4 > 1       Add 4 on both sides

3y > 1       Divide 3 on both sides to get "y" by itself

y>\frac{1}{3}

D.) -\frac{3}{4} +6y > \frac{1}{4}       Isolate "y", add 3/4 on both sides

6y > 1     Divide 6 on both sides to get "y" by itseld

y>\frac{1}{6}

7 0
3 years ago
What is the missing angle?
murzikaleks [220]

a straight line = 180 degrees

 so the missing angle is 180 - 130 = 50 degrees

8 0
3 years ago
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