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3 years ago

4x3 (4x-8) 6 (-8 x+3) How do u solve the problem

Mathematics

1 answer:

White raven [17]3 years ago

4 0

I'm not sure if that's what you meant.

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Solve for x: 3x − 2 > 5x + 10.

Assoli18 [71]

Answer: hopefully this is what you need

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

Inequality form:

X < -6

Interval Notation:

(-♾, -6)

8 0

3 years ago

Can you please simplify (x^2)^5 ?

Sloan [31]

(x^2)^5 = x^(2*5) = x^10

8 0

3 years ago

Is the function given by, (Picture Provided)

Sever21 [200]

Answer:

a. yes

Step-by-step explanation:

The given function is

$f(x)=\left \{ {{3x-2,if\:x\le3} \atop {10-x,if\:x\:>\:3}} \right.$

$f(3)=\left \{ {{3(3)-2,if\:x\le3} \atop {10-3,if\:x\:>\:3}} \right.=7$

$\lim_{x \to 3^-} f(x)=3(3)-2=7$

$\lim_{x \to 3^+} f(x)=10-3=7$

$\Rightarrow \lim_{x \to 3} f(x)=7$

Since;

$\lim_{x \to 3} f(x)=f(3)$

The function is continuous at $x=3$

7 0

3 years ago

A chocolate company produces 2 types of chocolate: type a and type b. The company selects 25 random packages of each type to che

Taya2010 [7]

Answer:

80 chocolates type 'a' and 240 chocolates type 'b'

Step-by-step explanation:

The company selects 25 random packages of each type from which one of type a has an incorrect weight and three of type b have an incorrect weight.

We can say that $\frac{1}{25}=0.04$ = 4% of the chocolates type 'a' have an incorrect weight while $\frac{3}{25}=0.12$ = 12% of chocolates type 'b' have an incorrect weight.

Now, if the company selects 2000 random chocolates of each type we ca predict that the amount of chocolates that have an incorrect wright will be:

Type 'a': 2000x0.04 = 80

Type 'b': 2000x0.12 = 240

So the company is expected to find 80 chocolates type 'a' and 240 chocolates type 'b' that have incorrect weight.

7 0

4 years ago

Math Help? Please! : )

Ray Of Light [21]

1/2 is the probability, or 50%. There are 4 teams of S and B n the Y, 2 are B. so its 2/4

5 0

3 years ago