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xenn [34]
3 years ago
9

Please HELP!!!

Mathematics
1 answer:
anygoal [31]3 years ago
5 0
I think the answer is (3,3) I’m sorry if it’s wrong
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Need help... Fast! will give brainiest <br> What is the solution set of x/4 &lt; 9/x ?
Ivenika [448]

Answer:

Option (4)

Step-by-step explanation:

Given inequality is,

\frac{x}{4}\leq \frac{9}{x}

\frac{x}{4}-\frac{9}{x}\leq 0

\frac{x^{2}-36}{4x}\leq 0

x ≥ 0 [For x = 0, fraction is not defined]

(x - 6)(x + 6) ≤ 0

x - 6 ≤ 0

x ≤ 6

Or

x + 6 ≤ 0

x ≤ -6

Therefore, solution set is (-∞, -6] ∪ (0, 6]

By plotting the solution area on a number line,

Option (4) will be the answer.

7 0
3 years ago
Write the product in its simplest form c^7×(-3c)​
sveticcg [70]

Answer:

-3c^8

Step-by-step explanation:

3 0
3 years ago
Find their total volume if x=10 feet, y=12 feet, and z= 7 feet
saveliy_v [14]
The volume would be 840 feet
4 0
3 years ago
Which is a zero of the quadratic function f(x) = 16x?? + 32x - 9?
STALIN [3.7K]

Answer:

f(x) = 16x2 + 32x - 9

 

To find the "zeros", set f(x) = 0:

 

    0 = 16x2 + 32x - 9

 

The equation is solvable by factoring or by using the quadratic formula.  The factors (use the "ac" method) are:

 

    0 = (4x-1)(4x+9)

 

    x = 1/4 = 0.25 and x = -9/4 = -2.25

Step-by-step explanation:

6 0
3 years ago
Classify each polynomial by degree and number of terms. 1. Quartic monomial v 5x? - 7x + 9 2. Cubic binomial 2x3 + 3 &lt; 3. Qua
ElenaW [278]

The degree of a polynomial is categorized by its highest power of its leading variable.

<h3>Degree and number of terms of a polynomial</h3>

Given the following polynomial functions, we are to classify them based on their degrees and the number of terms

For the polynomial function 5x^4 - 7x + 9

The degree is 4 being quartic and there are three terms in the expression

For the polynomial  2x3 + 3 < 3, the degree is 3 (cubic) with three terms in the expression.

For a Quintic quadnomial​, the degree of the polynomial is 5.

Learn more on polynomials here: brainly.com/question/4142886

4 0
2 years ago
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