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kaheart [24]
2 years ago
15

Based on what you know now, how are linear and exponential functions alike?

Mathematics
1 answer:
Ainat [17]2 years ago
5 0

Answer:

They're similar in that they both have to maintain a steady rate of rise as they grow. While graphing, you can't adjust the slope or exponent after traveling up a graph.

Step-by-step explanation:

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Help me out thank u !!!!!
AlekseyPX

Answer:

<u>6.2 = Altitude</u>

Step-by-step explanation:

Hypotenuse = 8

Base = 5

Altitude = x

Hypotenuse² = Base² + Altitude²

8² = 5² + Altitude²

64 = 25 + Altitude²

64 - 25 = Altitude²

39 = Altitude²

√39 = Altitude

<u>6.2 = Altitude</u>

6 0
3 years ago
100 points plz help fasttt
prohojiy [21]

Answer:

Step-by-step explanation:

Complementary angles are 2 angles when added together = equal 90 * degrees.

The verticle angles theorem states angles opposite each other are the same degree*.

30* degrees is opposite angle 2 .

So angle 2 = 30* too

angle 2 and 3 together +equal 90* degree.

90 - 30 (angle 2) = 60*

Angle 3 is 60*

3 0
2 years ago
Read 2 more answers
A system of linear equations contains two equations with negative reciprocal
aleksandrvk [35]

let's recall that perpendicular lines have negative reciprocal slopes, so say for a line with a slope of a/b

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{a}{b}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{b}{a}}\qquad \stackrel{negative~reciprocal}{-\cfrac{b}{a}}}

a perpendicular one will be -b/a.

what does that mean?  well, that the lines are perpendicular of course, they meet at a 90° angle, and two straight lines meeting at 90° can only meet once.

since a solution to a system of equations is where the graph meet, and these two meet  perpendicularly once, then that means only one solution.

4 0
3 years ago
A consumer electronics company is comparing the brightness of two different types of picture tubes for use in its television set
maw [93]

Answer:

________________________________________________

5 0
3 years ago
Can you break apart 5/6 into three fraction with different numerators but the same denominators
dexar [7]

\bf \cfrac{5}{6}\implies \cfrac{1+7-3}{6}\implies \stackrel{\textit{distributing the denominator}}{\cfrac{1}{6}+\cfrac{7}{6}-\cfrac{3}{6}}

7 0
3 years ago
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