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3 years ago

Let log P/N=8 and log M/N=5 What is the relationship between P and M?

1 answer:

8 0

Want to use some of the algebra rules of log's:

log(a)-log(b)= log(a/b),

so log(P/N) - log(M/N) = log( P/N:M/N) = log(P/M),

Then: 8-5 = log(P/M), log(P/M)=3.

If log here means natural logarithm (base e), then P/M = e^3,

If log here means basis 10, decimal logarithm, then P/M = 10^3 = 1000.

You might be interested in

A line passes through the point (4,7) and has a slope of -5/2.

pshichka [43]

Answer:

$y=-\frac{5}{2} x+17$

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

Whose textbook reaches the ground first and by how many seconds?

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

3 0

3 years ago

Find the equation of the line that passes through the points (-5,7) and (2,3)

Olegator [25]

Answer:

$y=-\frac{4}{7}x+4\frac{1}{7}$

Step-by-step explanation:

So, in order to solve this problem, I started off by drawing it out. On my graph that I have attached below, I first started out by locating the points (-5,7) and (2,3). Now, this is an optional step, but I highly encourage practicing your graphing skills by solving this problem on graph paper as well. Next, I connected the two points that I just graphed. This is the line that passes through (-5,7) and (2,3).

Now, here is where the actual solving starts. If you haven't already been taught this yet, I will introduce it to you now. I am going to find the equation of this line by filling in what I know in the equation y=mx+b, where m= the slope of the line, and b= y intercept.

Slope of the line: m= $\frac{y_{1} - y_{2} }{x_{1} - x_{2} } = \frac{7-3}{-5-2} = \frac{4}{-7}= -\frac{4}{7}$

If you haven't been taught how to find the slope of a line I recommend you find out.

Substitute the slope into the equation.

$y=-\frac{4}{7} x+b\\$

Now, we will solve for the 'b,' or y intercept.

We already have x and y values to use: (-5,7) or (2,3). I'll use x=2 and y=3 to solve for the y intercept.

$y=-\frac{4}{7} x+b\\\\3=-\frac{4}{7} *2+b\\\\3=-\frac{8}{7} +b\\b=3+\frac{8}{7} \\b=\frac{21}{7}+\frac{8}{7}=\frac{29}{7} =4\frac{1}{7} \\b=4\frac{1}{7}$

Last step: substitute the slope and y intercept into y=mx+b.

$y=mx+b\\y=-\frac{4}{7}x+4\frac{1}{7}$

That is the answer to this problem.

I hope this helps.

3 0

3 years ago

3. if kx² + 2x + k = -kx have equal roots, find the values of k.

crimeas [40]

Answer:

k = - $\frac{2}{3}$ , k = 2

Step-by-step explanation:

Using the discriminant Δ = b² - 4ac

The condition for equal roots is b² - 4ac = 0

Given

kx² + 2x + k = - kx ( add kx to both sides )

kx² + 2x + kx + k = 0 , that is

kx² + (2 + k)x + k = 0 ← in standard form

with a = k, b = 2 + k and c = k , thus

(2 + k)² - 4k² = 0 ← expand and simplify left side

4 + 4k + k² - 4k² = 0

- 3k² + 4k + 4 = 0 ( multiply through by - 1 )

3k² - 4k - 4 = 0 ← in standard form

(3k + 2)(k - 2) = 0 ← in factored form

Equate each factor to zero and solve for k

3k + 2 = 0 ⇒ 3k = - 2 ⇒ k = - $\frac{2}{3}$

k - 2 = 0 ⇒ k = 2

5 0

3 years ago

-4(j+k+g) plz help using box method

disa [49]

Answer:

-4j-4k-4g

Step-by-step explanation:

8 0

3 years ago