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yuradex [85]
3 years ago
9

B2≤4 What is the solution to the inequality? b≥8 b≤8 b≤2 b≥2

Mathematics
2 answers:
Allisa [31]3 years ago
8 0
The solution is the third one:
b≤2
telo118 [61]3 years ago
8 0
Should be b is less than or equal to 2.

b2 < 4 divide both sides by 2 gives you your answer. Just be sure to put the line under to show the equals b < 2
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Image on straight line graphs- please help ASAP
MissTica
Plug in (7, 12) for x and y in the equation.

12 = 2(7) + 1
12 = 14 + 1
12 ≠ 15 

Therefore, (7, 12) is not on the straight line equation. 
6 0
3 years ago
What is 7(x + 6) solve it
Elis [28]

Answer:

49

Step-by-step explanation:

First you must do x times 6, x is really just 1. After you do that, you times it by 7.

8 0
3 years ago
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JOD, hele he earns $8 per hour. His employer always rounds up to the nearest quarter of an hour. For example, if
Nesterboy [21]

Answer:

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2 years ago
Type the correct answer in the box. Use a comma to separate the x- and y-coordinates of each point. The coordinates of the point
MAVERICK [17]

Answer:

On a unit circle, the point that corresponds to an angle of 0^{\circ} is at position (1, \, 0).

The point that corresponds to an angle of 90^{\circ} is at position (0, \, 1).

Step-by-step explanation:

On a cartesian plane, a unit circle is

  • a circle of radius 1,
  • centered at the origin (0, \, 0).

The circle crosses the x- and y-axis at four points:

  • (1, \, 0),
  • (0, \, 1),
  • (-1,\, 0), and
  • (0,\, -1).

Join a point on the circle with the origin using a segment. The "angle" here likely refers to the counter-clockwise angle between the positive x-axis and that segment.

When the angle is equal to 0^\circ, the segment overlaps with the positive x-axis. The point is on both the circle and the positive x-axis. Its coordinates would be (1, \, 0).

To locate the point with a 90^{\circ} angle, rotate the 0^\circ segment counter-clockwise by 90^{\circ}. The segment would land on the positive y-axis. In other words, the 90^{\circ}-point would be at the intersection of the positive y-axis and the circle. Its coordinates would be (0, \, 1).

4 0
3 years ago
Find the equation of a line that passes through (-5,4) and us parallel to y=-6/7x+3​
irina [24]

Given:

The given equation of line is

y=-\dfrac{6}{7}x+3

To find:

The equation of line that passes through (-5,4) and is parallel to the given line.

Solution:

Slope intercept form of a line is

y=mx+b    ...(i)

where, m is slope and b is y-intercept.

We have,

y=-\dfrac{6}{7}x+3      ...(ii)

On comparing (i) and (ii), we get

m=-\dfrac{6}{7}

Slope of given line is m=-\dfrac{6}{7}.

Slope of parallel lines are same. So, slope of parallel line is m=-\dfrac{6}{7}.

The required line passes through (-5,4) with slope m=-\dfrac{6}{7}. So, the equation of line is

y-y_1=m(x-x_1)

y-4=-\dfrac{6}{7}(x-(-5))

y-4=-\dfrac{6}{7}(x+5)

y-4=-\dfrac{6}{7}x-\dfrac{30}{7}

Adding 4 on both sides, we get

y=-\dfrac{6}{7}x-\dfrac{30}{7}+4

y=-\dfrac{6}{7}x+\dfrac{-30+28}{7}

y=-\dfrac{6}{7}x-\dfrac{2}{7}

Therefore, the correct option is a.

8 0
3 years ago
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