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Rudik [331]
3 years ago
8

Could you help me with this question?

Mathematics
1 answer:
Dafna11 [192]3 years ago
7 0

Answer:

2nd

Step-by-step explanation:

5=a^x

5/a

a^x/a

take a to the top

a^x.a^-¹

a^x-1

hope u understand

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|x-1| if x&gt;1<br> how do you simplify this
JulsSmile [24]

Use the definition of absolute value.

|x| = \begin{cases} x & \text{if } x \ge0 \\ -x & \text{if } x < 0 \end{cases}

The absolute value is always non-negative. So if x is already non-negative, |x|=x is unchanged. But if x is negative, then |x|=-x because multiplying x by -1 makes it positive.

Now, if x>1, then by the definition of absolute value,

x > 1 \implies x - 1 > 0 \implies \boxed{|x - 1| = x - 1}

4 0
1 year ago
Hi if someone can answer the top one I’ll mark u brainliest thanks
Kryger [21]

Answer:

the answer is a

Step-by-step explanation:

i got corrected :)

6 0
3 years ago
Read 2 more answers
An airplane leaves an airport at 9:00 p.M. With a heading of 270 and a speed of 610 mph. At 10:00 p.M. The pilot changes the he
timama [110]

Answer:

2330.51 miles

Step-by-step explanation:

Given that the speed of the airplane = 610 mph.

The airplane leaves an airport at 9:00 P.M. with a heading of 270 degrees and at 10:00 P.M., the pilot changes the heading to 310 degrees.

So, for 1 hour the plane is heading at 270 degrees and for 3 hours, from 10:00 p.m to 1:00 a.m, the plane is heading at 310 degrees as shown in the figure.

As, distance = time x speed, so

The distance covered at 270 degrees, d_1 = 1\times610=610 miles.

The distance covered at 310 degrees, d_2 = 3\times610=1830 miles.

Total distance covered, d, is the magnitude of the sum of vectors \vec{d_1} and \vec{d_2} as shown in the figure.

The angle between the vectors \vec{d_1} and \vec{d_2}, \theta=310-270=40 degree.

Magnitude of sum of the vectors \vec{d_1} and \vec{d_2},

d= \sqrt{|\vec{d_1}|^2+|\vec{d_2}|^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\theta} \\\\\Rightarrow d=\sqrt{610^2+1830^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\(40^{\circ})}

\Rightarrow d=2330.51 miles

Hence, at 1:00 a.m, the airplane is at a distance of 2330.51 miles from the airport.

4 0
3 years ago
Triangle abc is such that ab=4 and ac=8 if m is the midpoint of bc and am=3, what is the length of bc?
zavuch27 [327]
This is an interesting question. I chose to tackle it using the Law of Cosines.
  AC² = AB² + BC² - 2·AB·BC·cos(B)
  AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
  AC² - 2·AM² = -AB² + BC² - 2·MB²

We know that MB = BC/2. When we substitute the given information, we have
  8² - 2·3² = -4² + BC² - BC²/2
  124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
  2√31 = BC ≈ 11.1355

4 0
3 years ago
A 5000 seat theater has tickets for sale at 26 and 40. How many tickets should be sold at each price for a sellout performance t
slavikrds [6]

Answer:

The number of tickets for sale at $26 should be 3300

The number of tickets for sale at $40 should be 1700

Step-by-step explanation:

Use 2 equations to represent the modifiers within the problem:

5000 = a + b \\ 153800 = 26a + 40b

Now you want to find the point at which the variables are changed to make both equations correct, this can be done by graphing and finding the intersection of both lines.

5000 = 3300 + 1700 \\ 153800 = 26(3300) + 40(1700)

5 0
3 years ago
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