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3 years ago

Find the solution(s) for the equation below: |3x – 2| - 5 = -1

Mathematics

2 answers:

ira [324]3 years ago

7 0

Answer:

It would be D.x = -2/3 or x =2

BigorU [14]3 years ago

3 0

Answer:

Step-by-step explanation:

(3x-2) - 5 = -1

⇒(3x-2)-4=0

⇒3x-6=0

⇒3x=6

⇒x=6/3

⇒x=2

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Please answer the following question. Also please answer if you actually know it.

Iteru [2.4K]

Given : f(x)= 3|x-2| -5

f(x) is translated 3 units down and 4 units to the left

If any function is translated down then we subtract the units at the end

If any function is translated left then we add the units with x inside the absolute sign

f(x)= 3|x-2| -5

f(x) is translated 3 units down

subtract 3 at the end, so f(x) becomes

f(x)= 3|x-2| -5 -3

f(x) is translated 4 units to the left

Add 4 with x inside the absolute sign, f(x) becomes

f(x)= 3|x-2 + 4| -5 -3

We simplify it and replace f(x) by g(x)

g(x) = 3|x + 2| - 8

a= 3, h = -2 , k = -8

5 0

3 years ago

What is the perimeter of the rectangle shown below? apex algebra 1 sem 1 unit 6

Allushta [10]

Great job! you didn't include the rectangle! no one can answer this question! you're the best!

7 0

3 years ago

Read 2 more answers

: A new bear population that begins with 150 bears in 2000 decreases at a rate of 20% per year.

d1i1m1o1n [39]

Answer:

2 bears in 2020.

Step-by-step explanation:

We have been given that a new bear population that begins with 150 bears in 2000 decreases at a rate of 20% per year.

We will use exponential decay formula to solve our given problem as:

$y=a\cdot (1-r)^x$ , where,

y = Final quantity,

a = Initial value,

r = Decay rate in decimal form,

x = Time

$20\%=\frac{20}{100}=0.20$

Upon substituting our given values in above formula, we will get:

$y=150(1-0.20)^x$

$y=150(0.80)^x$ , where x corresponds to year 2000.

To find the population in 2020, we will substitute $x=20$ in our equation as:

$y=150(0.80)^{20}$

$y=150(0.011529215046)$

$y=1.7293822569$

$y\approx 2$

Therefore, 2 bears are there predicted to be in 2020.

Since population is decreasing so population is best described as exponential decay.

8 0

3 years ago

The distance of a golf ball from the hole can be represented by the right side of a parabola with vertex (-1,8). The ball reache

ddd [48]

Answer:

Let's suppose that the hole is at y = 0m, where x is the time variable.

we know that:

The vertex is (-1s, 8m).

(i suppose x in seconds and y in meters)

At x = 1s, the ball reaches the hole, so we also have the point:

(1s, 0m).

Remember that the vertex of a quadratic equation y = a*x^2 + b*x + c is at:

x = -b/2a,

then we have:

-1 = -b/2a.

Then we have tree equations:

8m = a*(-1s)^2 + b*-1s + c

0m = a*(1s)^2 + b*1s + c

-1s = -b/2a.

First we should isolate one variable in the third equation, and then replace it in one of the other two:

1s*2a = b.

So we can replace b in the first two equations bi 1s*2a.

8m = a*1s^2 - 1s*2a*1s + c

0m = a*1s^2 + 1s*2a*1s + c

We can simplify both equations and get:

8m = a*( 1s^2 - 2s^2) + c = -a*1s^2 + c.

0m = a*(1s^2 + 2s^2) + c = a*3s^2 + c.

Easily we can isolate c in the second equation and then replace it into the first equation:

c = -a*3s^2

The first equation becomes:

8m = -a*1s^2 - a*3s^2 = -a*4s^2

a = 8m/-4s^2 = -2m/s^2.

Now with a, we can find the values of c and b.

c = -a*3s^2 = -(-2m/s^2)*3s^2 = 6m.

b = 1s*2a = 1s*(-2m/s^2) = -2m/s.

Then the equation is:

y = (-2m/s^2)*t^2 + (-2m/s)*t + 6m

5 0

2 years ago

ASAP what is the value of x

Nataliya [291]

2x+2= 3x-52

Solve for x

You get x to be 54

8 0

3 years ago