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nataly862011 [7]
2 years ago
15

What scale factor was applied to the first rectangle to get the resulting image?

Mathematics
2 answers:
romanna [79]2 years ago
4 0

Answer:

2.5x

Step-by-step explanation:

7.5 / 3 = 2.5

scoray [572]2 years ago
4 0

Answer:

2.5

Step-by-step explanation:

now to find the scale factor we do

7.5/3

=2.5

so the rectangle is increased by the scale factor of 2.5

<em>I work hard on my answers</em>

<em>it would be very appreciated if you award me with a brainliest</em>

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Find all solutions to the equation (x² + 5x + 6)(x² − 3x − 4) = 0.
Soloha48 [4]

Answer:

The solutions are:

x=-2,\:x=-3,\:x=-1,\:x=4

Step-by-step explanation:

To find the solutions to the equation \left(x^2+5x+6\right)\left(x^2-3x-4\right)=0 you need to:

  • Factor \left(x^2+5x+6\right)

Break the expression into groups

x^2+5x+6=\left(x^2+2x\right)+\left(3x+6\right)

Factor out x from (x^2+2x)

x^2+2x=x\left(x+2\right)

Factor out 3 from 3x+6

3x+6=3\left(x+2\right)

x^2+5x+6=x\left(x+2\right)+3\left(x+2\right)\\\\\mathrm{Factor\:out\:common\:term\:}x+2\\\\x^2+5x+6=\left(x+2\right)\left(x+3\right)

  • Factor \left(x^2-3x-4\right)

x^2-3x-4=\left(x^2+x\right)+\left(-4x-4\right)\\\\x^2-3x-4=x\left(x+1\right)-4\left(x+1\right)\\\\x^2-3x-4=\left(x+1\right)\left(x-4\right)

Therefore

\left(x^2+5x+6\right)\left(x^2-3x-4\right)=\left(x+2\right)\left(x+3\right)\left(x+1\right)\left(x-4\right)=0

Using the Zero Factor Theorem:

5 0
4 years ago
Will mark brainlist if its right
pickupchik [31]

Answer: S<\-2

Hope i was some help.

Step-by-step explanation: sorry I can’t put the right sign my keybored won’t let me but you know what it stands for.

3s+6s</-5(s+2)

start off with -5(s+2) you multiply -5 with s and it is -5s then you do the same for +2 take -5 and multiply by +2 pos times a neg is always a neg so it’s -10.

3s+6<\-5s-10

So you have to get rid of the -5s so your going to do the inverse which is add 5 to -5 and it will get rid of -5s but then you have to add 5 to 3s and that will be 8s.

8s+6</-10

Now your going to get rid of the +6 and to do that you do the inverse you subtract 6 and it will get rid of it now you have to do the same for -10 so you will subtract 6 to -10 and that will equal -16.

Now you have to divide!

8s/8 </ -16/8

And it should equal s <\-2

3 0
3 years ago
(Identify the Terms and Like Terms in the expression)<br> 10x+5+3x+1
UkoKoshka [18]
The like terms in this problem are the 10x and the 3x, as well as the 5 and 1. The 10x and 3x combine to get 13x, and the 5 and 1 add to make six.
6 0
3 years ago
Two thirds the volume
Fantom [35]

We can write this as :
1/3 x 2 or 2/1

4 0
3 years ago
Determine la ecuación general de la recta que pasa por los puntos (1,4); (- 2, - 5) y grafíquela.
Rudiy27

The equation of the line that passes through the points (1 , 4) and (-2, -5) is y = 3x + 1

<h3>Further explanation</h3>

Solving linear equation mean calculating the unknown variable from the equation.

Let the linear equation : y = mx + c

If we draw the above equation on Cartesian Coordinates , it will be a straight line with :

<em>m → gradient of the line</em>

<em>( 0 , c ) → y - intercept</em>

Gradient of the line could also be calculated from two arbitrary points on line ( x₁ , y₁ ) and ( x₂ , y₂ ) with the formula :

\large {\boxed {m = \frac{y_2 - y_1}{x_2 - x_1}} }

If point ( x₁ , y₁ ) is on the line with gradient m , the equation of the line will be :

\large {\boxed{y - y_1 = m ( x - x_1 )} }

Let us tackle the problem.

Let :

(1 , 4) → (x₁ , y₁)

(-2, -5) → (x₂ , y₂)

To find the straight line equation, the following formula can be used :

\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}

\frac{y - 4}{-5 - 4} = \frac{x - 1}{-2 - 1}

\frac{y - 4}{-9} = \frac{x - 1}{-3}

\frac{y - 4}{3} = \frac{x - 1}{1}

y - 4 = 3 ( x - 1 )

y = 3x - 3 + 4

\large {\boxed {y = 3x + 1} }

<h3>Learn more</h3>
  • Infinite Number of Solutions : brainly.com/question/5450548
  • System of Equations : brainly.com/question/1995493
  • System of Linear equations : brainly.com/question/3291576

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Linear Equations

Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point

4 0
4 years ago
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