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strojnjashka [21]
4 years ago
15

What is the x-intercept for this linear equation? 6x-y=9

Mathematics
2 answers:
Katyanochek1 [597]4 years ago
7 0
The x intercept is:
(3/2,0)
Anastasy [175]4 years ago
3 0
The answer is (3,20)
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Find the mean, median, mode, and range of the data set,
ozzi

Answer:

C

Step-by-step explanation:

8 0
3 years ago
Lauren is reading a novel for English class. The data below shows the number of pages reads, y, in x hours.
strojnjashka [21]

Answer:

Choice B is correct

Step-by-step explanation:

The variable x represents the number of hours;

2, 4, 5, 7, 9

The least value is 2 while the maximum value is 9. The scale to be used must accommodate all these values. The scale can thus begin from 0 to 10. Choice A is incorrect since the values 7 and 9 will not be accommodated in the graph.

The y variable represents the number of pages read;

50, 100, 125, 175, 225

The least value is 50 while the maximum value is 225. The scale to be used must accommodate all these values. The scale can thus begin from 0 to 250. Both choices C and D are incorrect since none of the values will be accommodated in the graph.

7 0
3 years ago
Read 2 more answers
Eight less than the product of 2 and a number equals 5
Novay_Z [31]

Answer:

6.5

Step-by-step explanation:

I'm not sure if its right but:

(2 × x) - 8 = 5

2x = 13

x = 6.5

6 0
4 years ago
1) a -4 - b; use a = 6, and b = 1
Temka [501]

Answer:

-23

Step-by-step explanation:

6×-4 -1.... (6)(-4)-1

-24-1.....-24-1

-23

4 0
3 years ago
Which of the following is a solution of x^2 + 2x + 4?
Kaylis [27]

Answer:

\displaystyle x_1=-1+\sqrt{3}i

\displaystyle x_2=-1-\sqrt{3}i

Step-by-step explanation:

<u>Second-Degree Equation</u>

The second-degree equation or quadratic equation has the general form

ax^2+bx+c=0

where a is non-zero.

There are many methods to solve the equation, one of the most-used is by using the solver formula:

\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

The equation of the question has the values: a=1, b=2, c=4, thus the values of x are

\displaystyle x=\frac{-2\pm \sqrt{2^2-4\cdot 1\cdot 4}}{2\cdot 1}

\displaystyle x=\frac{-2\pm \sqrt{-12}}{2}

Since the square root has a negative argument, both solutions for x are imaginary or complex. Simplifying the radical

\displaystyle x=\frac{-2\pm 2\sqrt{-3}}{2}=-1\pm\sqrt{3}i

The solutions are

\displaystyle x_1=-1+\sqrt{3}i

\displaystyle x_2=-1-\sqrt{3}i

7 0
3 years ago
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