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3 years ago

Solve 4^x+2 = 2^x-1. What are the exact and approximate solutions?

Mathematics

2 answers:

Brut [27]3 years ago

7 0

Answer:

C. x = -5 ln 2/ ln 2 = -5

Step-by-step explanation:

Juliette [100K]3 years ago

6 0

Answer:

no solution

Step-by-step explanation:

examples: 4^1 + 2 = 2^ 1 -1 and the answer is 6 = 1 which is not true. So its gonna be hard for them to have a solution.

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Find an equation for the perpendicular bisector of the line segment whose endpoints are ( − 1 , − 1 ) (−1,−1) and ( 9 , 7 ) (9,7

Hunter-Best [27]

Answer:

Place the compass at one end of line segment.

Adjust the compass to slightly longer than half the line segment length.

Draw arcs above and below the line.

Keeping the same compass width, draw arcs from other end of line.

Place ruler where the arcs cross, and draw the line segment.

3 0

3 years ago

Read 2 more answers

Eric was rock climbing. At one point, he stopped and climbed straight down 2\dfrac{1}{2}2

nikitadnepr [17]

The equation that matches the given situation is $-2\frac{1}{2}+6\frac{3}{4}=?$ . So, after two moves, Eric's elevation changed $4\frac{1}{4}$ meters above.

We know that in mathematics, subtraction means removing something from a group or a number of things. What is left in the group gets smaller when we subtract. The minuend is the first element we use. The subtrahend is the part that is being removed. The difference is the portion that remains after subtraction.

Assume that "negative" means climbing down and "positive" means climbing up. We must locate the elevation change in this area.

Given that Eric climbed straight down $2\frac{1}{2}$ meters. So we can write $-2\frac{1}{2}$ .

Again Eric climbed straight up $6\frac{3}{4}$ meters. So, we can write $6\frac{3}{4}$ .

Then the change = $-2\frac{1}{2}+6\frac{3}{4}$

= $-\frac{(2*2+1)}{2}+\frac{6*4+3}{4}$

= $-\frac{5}{2} +\frac{27}{4}$

= $\frac{-(5*2)+27}{4}$

= $\frac{-10+27}{4}$

= $\frac{17}{4}$

= $\frac{4*4+1}{4}$

= $4\frac{1}{4}$

Therefore, the equation that matches the given situation is $-2\frac{1}{2}+6\frac{3}{4}=?$ . So, after two moves, Eric's elevation changed $4\frac{1}{4}$ meters above.

Learn more about subtraction here -

brainly.com/question/24116578

#SPJ10

7 0

1 year ago

Find the midpoint of points A (5,-6) and B(2,0) graphically

netineya [11]

I think its (3.5, -3)

8 0

2 years ago

The functions below represent the amounts of water released from two reservoirs over x weeks:

solong [7]

Answer:

h(x) = -2x^2 -x +3
Reservoir A releases the same amount of water as Reservoir B over 1 week

Step-by-step explanation:

Given:

f(x) = x^2 -7x +5

g(x) = 3x^2 -6x +2

h(x) = f(x) -g(x)

Find:

h(x)

h(1)

Solution:

The expression for h(x) is found by evaluating its definition:

h(x) = (x^2 -7x +5) -(3x^2 -6x +2)

h(x) = x^2(1 -3) +x(-7 -(-6)) +(5 -2)

h(x) = -2x^2 -x +3

Then h(1) is found by substituting 1 for x:

h(1) = -2(1^2) -(1) +3 = -2 -1 +3

h(1) = 0 . . . . difference in release amounts after 1 week is 0

The true statements are ...

h(x) = -2x^2 -x +3
Reservoir A releases the same amount of water as Reservoir B over 1 week

6 0

3 years ago

EXPLAIN why we placed the value of x= 4/3( the minimum value) into the equ of gradient(dy/dx) [in the answer, marking scheme att

aliina [53]

$y=x(x-2)^2$
$\implies y'=(x-2)^2+2x(x-2)=3x^2-8x+4=(3x-2)(x-2)=0$
$\implies x=\dfrac23,x=2$

are the critical points, and judging by the picture alone, you must have $b=\dfrac23$ and $a=2$ . (You might want to verify with the derivative test in case that's expected.)

Then the shaded region has area

$\displaystyle\int_0^2x(x-2)^2\,\mathrm dx=\dfrac43$

I'll leave the details to you.

Now, for part (iv), you're asked to find the minimum of $\dfrac{\mathrm dy}{\mathrm dx}=y'$ , which entails first finding the second derivative:

$y'=3x^2-8x+4$
$\implies y''=6x-8$

setting equal to 0 and finding the critical point:

$6x-8=0\implies x=\dfrac86=\dfrac43$

This is to say the minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ *occurs when $x=\dfrac43$ *, but this is not necessarily the same as saying that $\dfrac43$ is the actual minimum value.

The minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ is obtained by evaluating the derivative at this critical point:

$m=\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=4/3}=3\left(\dfrac43\right)^2-8\left(\dfrac43\right)+4=-\dfrac43$

4 0

3 years ago

Solve 4^x+2 = 2^x-1. What are the exact and approximate solutions?​

Solve 4^x+2 = 2^x-1. What are the exact and approximate solutions?