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3 years ago

Which graph shows a proportional relationship between x and y??

Mathematics

1 answer:

Lena [83]3 years ago

6 0

Answer:B

Step-by-step explanation:

B is has a proportional x and y value because they're increasing at the same rate which is 0

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Solve −12 − 2x < −3x + 2 < −2(x − 7). 14 over 5 < x < negative 12 over 5 −12 < x < 14 14 < x < −12 12 ov

zepelin [54]

$-12-2x < -3x+2 < -2(x-7)\\\\-12-2x < -3x+2 < -2x+14\ \ \ \ |+2x\\\\-12 < -x+2 < 14\ \ \ \ |-2\\\\-14 < -x < 12\ \ \ \ |\text{change the signs}\\\\14 > x > -12$

3 0

3 years ago

Please solve this problem. 2x + 34 = 56

maksim [4K]

Answer:

x = 11

Step-by-step explanation:

2x + 34 = 56

2x = 56 - 34

2x = 22

x = 22/2

x = 11

8 0

3 years ago

Read 2 more answers

If x =2 find the value of q in the equation 3x-4=x+q

Effectus [21]

Step-by-step explanation:

given x=2

and the equation

$3x - 4 = x + q$

to find the value of q , you need to make q the subject in the equation as shown below

q= 3x-x -4

q=2x-4

but x= 2

so just substitute the value of x in the equation for the q

$q = 2(2) - 4$

q=4-4

q=0

therefore the value of q is 0

5 0

3 years ago

Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).

Serjik [45]

Answer:

$\displaystyle f(x)=x^2+2x+2$

Step-by-step explanation:

System Of Linear Equations

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

$\displaystyle f(x)=ax^2+bx+c$

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

$\displaystyle f(-1)=a(-1)^2+b(-1)+c$

$\displaystyle f(-1)=a-b+c$

$\displaystyle a-b+c=1.....[eq\ 1]$

Now we use the second condition f(1)=5

$\displaystyle f(1)=a(1)^2+b(1)+c$

$\displaystyle f(1)=a+b+c$

$\displaystyle a+b+c=5.......[eq\ 2]$

Finally, we use the third condition f(2)=10

$\displaystyle f(2)=a(2)^2+b(2)+c$

$\displaystyle f(2)=4a+2b+c$

$\displaystyle 4a+2b+c=10....[eq\ 3]$

We put together eq 1, eq 2, and eq 3 to form the system

$\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.$

Adding the first two equations we have

$\displaystyle 2a+2c=6$

$\displaystyle a+c=3$

And also

$\displaystyle b=2$

Using the above equation and the value of b in the third equation, we have

$\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.$

Subtracting the first equation from the second

$\displaystyle 3a=3$

$\displaystyle a=1$

And therefore

$\displaystyle c=2$

Now we have all the values, the quadratic function is

$\displaystyle \boxed{f(x)=x^2+2x+2}$

6 0

3 years ago

If you run a marathon of 26 1/5 miles in a time of 5 4/5 how far do you run in 1 hour pls

alisha [4.7K]

First let's make both the miles and the hours ran improper fractions:

$26+\frac{1}{5}=\frac{131}{5}\\ 5+\frac{4}{5}=\frac{29}{5}$

So, to find our miles per hour, we have to divide the miles by the hours to get: $\frac{\frac{131}{5}\text{miles}}{\frac{29}{5}\text{hours}}=\frac{131\text{miles}}{29\text{hours}} \\ \text{ Now we have to divide the top and the bottom by 29 to get our miles per hour so: }\\ \frac{\frac{131}{29}}{\text{hour}} \approx \frac{4.52 \text{miles}}{\text{hour}}$

8 0

3 years ago