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2 years ago

11

What is the domain and range of this graph?

1 answer:

Mama L [17]2 years ago

6 0

Domain: (-1, 5)
Range: (-1, 3)

Explanation:
The domain covers the lowest and highest x value. You take the lowest number for the x and the highest number and that’s your domain. The range is covering y values, so you look at the lowest and highest point on your graph

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Solve the equation 3x+5y=15 for y

larisa86 [58]

<span> 3x+5y = 15
5y = -3x + 15
y = -3/5(x) + 3

hope it helps

</span>

7 0

3 years ago

50 points and brainliest to correct answer

Bumek [7]

Answer:

Its below!

Step-by-step explanation:

a=l*w

9=(3+x) (2+x)

9=6+3x+2x+x^2

9=6+5x+x^2

3=5x+x^2

Hope this helps!! :)

5 0

2 years ago

Global mean temperature was changing the most rapidly in 2005. How quickly was the temperature changing in 2005

AysviL [449]

The temperature was changing at a rate of 2.925 degrees Celsius in 2005

<h3>How to determine the rate in 2005</h3>

From the graph (see attachment), we have the following ordered pair

(x,y) = (5,14.625)

The above means that, the temperature in 2005 is 14.625 degrees Celsius

So, the rate in 2005 is:

$Rate = \frac{14.625}{5}$

$Rate = 2.925$

Hence, the temperature was changing at a rate of 2.925 degrees Celsius in 2005

Read more about average rates at:

brainly.com/question/8728504

4 0

2 years ago

What is the value of 12c5?<br><br> A. 95,040<br><br> B.792<br><br> C.120<br><br> D.60

nydimaria [60]

12x5 is 60 is it suppose to be 12 times 5

6 0

2 years ago

Read 2 more answers

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

or

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

or

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

or

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

2 years ago