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Effectus [21]
3 years ago
7

Graph the solution to the following inequality on the number line.x^2 +10x >or equal to-24​

Mathematics
1 answer:
zavuch27 [327]3 years ago
8 0

Answer:

x≤−6 or x≥−4

Step-by-step explanation:

We need to graph the inequality x^2+10x\ge -24

We can find the solution of the graph as follows :

x^2+10x+24\ge 0\\\\x^2+6x+4x+24\ge 0\\\\x(x+6)+4(x+6)\ge 0\\\\(x+4)(x+6)\ge 0\\\\x\le -6\ \text{and}\ x\ge -4

The attached figure shows the graph for the given inequality.

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6 0
3 years ago
What's the slope of the line passing through the points -1, 8 - 7, 3​
tatuchka [14]

Answer:

slope = 5/6

Step-by-step explanation:

Since you were given two points, you can use the point-slope formula to find the slope. The general equation looks like this:

y₁ - y₂ = m(x₁ - x₂)

In this formula, "m" represents the slope. To find the slope, plug the values from the two points into the equation. Make sure to put the values from the same point in the variable with the same number.

Point 1: (-1, 8)

Point 2: (-7, 3)

y₁ - y₂ = m(x₁ - x₂)                          <----- Original formula

8 - y₂ = m(-1 - x₂)                           <----- Plug in "x" and "y" values from Point 1

8 - 3 = m(-1 - (-7))                           <----- Plug in "x" and "y" values from Point 2

5 = m(-1 - (-7))                                <----- Simplify left side

5 = m(6)                                        <----- Simplify inside parentheses

5/6 = m                                       <----- Divide both sides by 6

5 0
2 years ago
Given the inequality of x &lt; 5, select all the
diamong [38]
Nxnxnxbbxnxndnnsnsn nsnsnsndnnsndnndndnsndnndndnndnd
7 0
3 years ago
Approximately what portion of the box is shaded blue?<br><br> A.2/3. B.9/10<br> C.3/5
vaieri [72.5K]
It is 9/10 as the others are too small
8 0
2 years ago
Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.
fredd [130]

\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}

\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

so the ODE is indeed exact and there is a solution of the form F(x,y)=C. We have

\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)

f'(y)=0\implies f(y)=C

\implies F(x,y)=x^2y^3+\ln(x+y^2)=C

With y(1)=2, we have

8+\ln9=C

so

\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

8 0
3 years ago
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