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galina1969 [7]
4 years ago
13

Select the graph that would represent the best presentation of the solution set for Ixl <5.

Mathematics
2 answers:
sattari [20]4 years ago
6 0

Answer:

< = region below graph

> = region above graph

Lisa [10]4 years ago
3 0

< = region below graph

> = region above graph

Step-by-step explanation:

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During a two-hour period, the temperature in a city dropped from a high of 51°F to a low of −11°F. What was the range of the tem
Likurg_2 [28]

you have to take 51 plus 11 and add it up. then your answer should be 62 degrees different.

6 0
3 years ago
Find the Inverse of
Mekhanik [1.2K]

Answer:

y = 1/2x + 3

O f(x)= 1/2x+3

Step-by-step explanation:

f(x) = 2x - 6

y = 2x - 6

x = 2y - 6

2y = x + 6

y = (x + 6)/2

y = 1/2x + 3

4 0
4 years ago
3x to the power of two minus x<br> Factor by gcf
algol13

Answer:

After factorizing the given expression we get the value as x(3x-1).

Step-by-step explanation:

Given:

3x^2-x

We need to factorize the given expression using GCF.

Solution:

3x^2-x

Now GCF means Greatest common factor.

From the given 2 numbers we need to find the greatest common factor.

3\times x\times x- 1 \times x

In the given expression GCF is 'x'.

Hence we can say that;

x(3x-1)

Hence After factorizing the given expression we get the value as x(3x-1).

4 0
3 years ago
Find all of the zeros of the function f(x)=x^3-17x^2+81x-65
viktelen [127]
A graph shows the function can be factored as
  f(x) = (x -1)((x-8)² +1)

There is one real root and there are two complex roots. The latter can be found from the vertex form factor:
  (x - 8)² + 1 = 0
  (x - 8)² = -1
  x - 8 = ±√(-1) = ±i
  x = 8 ± i

The zeros of the function are {1, 8-i, 8+i}.

4 0
3 years ago
For the right triangle shown, which expression represents the length of CB?
Rudik [331]

Answer:

B) 6 tan(35°)

Step-by-step explanation:

from the right angle tan° = opposite / adjacent

therefore tan (35°) = (x) / 6

by multiplying 6 to both sides

tan (35°)  * 6 = (x) / 6 * 6

therefore (x) = 6 tan (35°)

5 0
3 years ago
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