If it has no stretch factor, then there will be no coefficient. Meaning this is a lot easier. So, we equation is

(x+6)^2 + 5

Hope this helps!
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8 0

2 years ago

Can someone please help me its urgent!!!!!!!!!!<br> 3x + 11 = 3x - 11

user100 [1]

Answer:

$\Large\boxed{\mathsf{\Rightarrow NO\quad SOLUTIONS}}$

Step-by-step explanation:

<h3><u>TO SOLVE:</u></h3>

Isolate by the x on one side of the equation.

<h3><u>SOLUTIONS:</u></h3>

First, subtract 3x from both sides.

$\displaystyle \Rightarrow \mathsf{3x+11-3x=3x-11-3x}}$

Solve.

$\displaystyle \mathsf{11=-11}$

Sides are NOT equal.

$\Large\boxed{\mathsf{\Rightarrow THERE \quad ARE \quad NO \quad SOLUTIONS}}}$

Therefore, there are no solutions.

6 0

3 years ago

Please help!!

ZanzabumX [31]

The two linear equations in two variable is:

12 x + 3 y = 40

7 x - 4 y = 38

(a) For a system of equations in two Variable

a x + by = c

p x + q y = r

It will have unique solution , when

$\frac{a}{p}\neq \frac{b}{q}\neq\frac{c}{r}$

As, you can see that in the two equation Provided above

$\frac{12}{7}\neq \frac{3}{-4}\neq \frac{40}{38}$

So, we can say the system of equation given here has unique solution.

(b). If point (2.5, -3.4) satisfies both the equations, then it will be solution of the system of equation, otherwise not.

1. 12 x+3 y=40

2. 7 x-4 y=38

Substituting , x= 2.5 , and y= -3.4 in equation (1) and (2),

L.H.S of Equation (1)= 1 2 × 2.5 + 3 × (-3.4)

= 30 -10.20

= 19.80≠ R.H.S that is 40.

Similarly, L H S of equation (2)= 7 × (2.5) - 4 × (-3.4)

= 17.5 +13.6

= 31.1≠R HS that is 38

So, you can Write with 100 % confidence that point (2.5, -3.4) is not a solution of this system of the equation.

5 0

2 years ago

rode three miles farther each day than the day before for a week. In total one hundred and twenty six miles were rode. How far w

aalyn [17]

126
7 x 3 = 21

126 + 21 = 147

7 0

2 years ago