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3 years ago

I need help please?!!!

Mathematics

2 answers:

Natalija [7]3 years ago

5 0

No it is not a solution!

Rina8888 [55]3 years ago

4 0

Answer:

Step-by-step explanation:

To find out if the ordered pair is a solution, just plug x and y into the equation. Is 5(1) + (3) equal to or less than -3? No, it equals 8, which is greater than -3. So the answer is No.

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2 years ago

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Which is the solution set to the inequality (4x-3)(2x-1)&gt=0?

Serjik [45]

(4x - 3)(2x - 1) ≥ 0

First, find the zeros:

4x - 3 = 0 2x - 1 = 0

x = $\frac{3}{4}$ x = $\frac{1}{2}$

Next, plot these points and choose test points on the outside and between the zeros:

←-------0------ $\frac{1}{2}$ ------ $\frac{5}{8}$ ------ $\frac{3}{4}$ ------1------→

Lastly, plug in the test points and look for a positive result (since it is greater than 0).

Test Point 0: [4(0) - 3][2(0) - 1] = ( - )( - ) = + THIS WORKS!

Test Point $\frac{5}{8}$ : [4( $\frac{5}{8}$ ) - 3][2( $\frac{5}{8}$ ) - 1] = ( - )( + ) = - <em>This does NOT work</em>

Test Point 1: [4(1) - 3][2(1) - 1] = ( + )( + ) = + THIS WORKS!

Answer: x ≤ $\frac{1}{2}$ or x ≥ $\frac{3}{4}$

Interval Notation: (-∞, $\frac{1}{2}$ ] U [ $\frac{3}{4}$ , ∞)

Graph: ←------ $\frac{1}{2}$ $\frac{3}{4}$ --------→

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3 years ago

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3 years ago

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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:

<u>System Of Linear Equations </u>

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