All categories

3 years ago

−3x>24 solve the inequality

Mathematics

2 answers:

Len [333]3 years ago

8 0

Answer:

x < -8

Step-by-step explanation:

−3x>24

Divide each side by negative 3, remembering to flip the inequality

-3x/-3 < 24/-3

x < -8

weeeeeb [17]3 years ago

3 0

Answer:

x<-8

Step-by-step explanation:

−3x>24

Multiply both sides by (-1) (reverse the inequality)

(-3x)(-1)<24(-1)

Simplify;

3x<-24

Divide both sides by 3

3x/3<-24/3

Simplify

3x/3<-24/3

Simplify; 3x/3

Simplify; -24/3

Divide the numbers;24/3=8

=-8

x<-8

Hope this helped!!!

You might be interested in

Can someone help ASAP this is missing and destroyed my grade

maria [59]

Answer:

Step-by-step explanation:

6ft plus 1 3/4 is 7 3/4

Sorry for minimum explanation but I hope it helped!

7 0

3 years ago

Read 2 more answers

Equation parallel to y = 3x + 2 and passing through (1, 1)

Kisachek [45]

Answer:

y = 3x - 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 3x + 2 ← is in slope- intercept form

with slope m = 3

Parallel lines have equal slopes, then

y = 3x + c ← is the partial equation

To find c substitute (1, 1) into the partial equation

1 = 3 + c ⇒ c = 1 - 3 = - 2

y = 3x - 2 ← equation of line

6 0

3 years ago

Which one of the following is NOT true about the expression shown

garri49 [273]

Answer:

A) The expression is a trinomial.

Step-by-step explanation:

If the degree is 2, the expression would be a binomial, not a trinomial. If this answer is correct, please make me Brainliest!

7 0

3 years ago

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

3 0

3 years ago

What is the starting point of this function

Finger [1]

the starting point of the function is 0

6 0

3 years ago