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2 years ago

Solution of the system of inequalities

Mathematics

1 answer:

Agata [3.3K]2 years ago

5 0

Answer:

Step-by-step explanation:

(0,-1) is not part of the solution because:

It is not in the solution region and

the bottom line is not inclusive so it does not count.

Any points in the shaded region are solutions. Since (0,-1) is not in the shaded region it is not a solution.

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Describe and correct the error below.

Sergeeva-Olga [200]

Cos ( 390)
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6 0

2 years ago

Factor f(x) = 15x^3 - 15x^2 - 90x completely and determine the exact value(s) of the zero(s) and enter them as a comma separated

Illusion [34]

Answer:

$x=-2,0,3$

Step-by-step explanation:

We have been given a function $f(x)=15x^3-15x^2-90x$ . We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

$15x^3-15x^2-90x=0$

Now, we will factor our equation. We can see that all terms of our equation a common factor that is $15x$ .

Upon factoring out $15x$ , we will get:

$15x(x^2-x-6)=0$

Now, we will split the middle term of our equation into parts, whose sum is $-1$ and whose product is $-6$ . We know such two numbers are $-3\text{ and }2$ .

$15x(x^2-3x+2x-6)=0$

$15x((x^2-3x)+(2x-6))=0$

$15x(x(x-3)+2(x-3))=0$

$15x(x-3)(x+2)=0$

Now, we will use zero product property to find the zeros of our given function.

$15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0$

$15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0$

$\frac{15x}{15}=\frac{0}{15}\text{ (or) }x-3=0\text{ (or) }x+2=0$

$x=0\text{ (or) }x=3\text{ (or) }x=-2$

Therefore, the zeros of our given function are $x=-2,0,3$ .

7 0

3 years ago

Help asap I will give u huge amount of brainly points

Minchanka [31]

Answer:

$\boxed{\sf -\dfrac{5}{8}}$

Use BODMAS rule,

where:

brackets
order
division
multiplication
addition
subtraction

Given expression:

$\rightarrow \sf -\dfrac{1}{5} \times [\:4-14\times\left(\dfrac{1}{4} \right)^2]$

simplify exponent

$\rightarrow \sf -\dfrac{1}{5} \times [\:4-14\times\left(\dfrac{1}{16} \right)]$

$\rightarrow \sf -\dfrac{1}{5} \times [\:4-\(\dfrac{14}{16}]$

$\rightarrow \sf -\dfrac{1}{5} \times [\:4-\(\dfrac{7}{8}]$

$\rightarrow \sf -\dfrac{1}{5} \times [\:\dfrac{32}{8} -\(\dfrac{7}{8}]$

$\rightarrow \sf -\dfrac{1}{5} \times [\:\dfrac{25}{8}]$

multiply fractions

$\rightarrow \sf -\dfrac{25}{40}$

$\rightarrow \sf -\dfrac{5}{8}$

8 0

1 year ago

Read 2 more answers

If P(3,2) and Q(7,10) are the endpoints of the diameter of a circle, what is the area of the circle in square units?

eimsori [14]

the diamiter is the two out side numbers minused by 3.14

4 0

3 years ago

A researcher computes r = 0.45 using a sample of 18 participants. Is this Pearson correlation coefficient significant for a two-

Eduardwww [97]

Answer with explanation:

Given : The computed r -value = 0.45

Sample size : n=18

Degree of freedom : $df=n-2=18-2=16$

Now, the critical value for Pearson correlation coefficient for a two-tailed test at a .05 level of significance will be :

$r_c=0.468$ ( by critical correlation coefficient table)

Since , $r>r_c$ i.e. 0.45>0.468 , then we say that his Pearson correlation coefficient is not significant for a two-tailed test at a .05 level of significance.

3 0

2 years ago