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3 years ago

Given log630 ≈ 1.898 and log62 ≈ 0.387, log615 =

2 answers:

6 0

$\log_630\approx1.898\\\\\log_62\approx0.387\\\\\log_615=\log_6\dfrac{30}{2}=\log_630-\log_62\approx1.898-0.387=1.511\\\\Used:\\\\\log_ab-\log_ac=\log_a\dfrac{b}{c}$

Oksi-84 [34.3K]3 years ago

4 0

Answer:

The value of $log_6{15}$ is 1.511.

Step-by-step explanation:

It is given that $log_6{30}\approx 1.898$ and $log_6{2}\approx 0.387$ .

We have to find the value of $log_6{15}$ .

It can be written as

$log_6{15}=log_6\frac{30}{2}$

Using property of lo, we get

$log_6{15}=log_630-log_62$ $log_m\frac{a}{b}=log_ma-log_mb$

Substitute $log_6{30}\approx 1.898$ and $log_6{2}\approx 0.387$ in the above equation.

$log_6{15}=1.898-0.387$

$log_6{15}=1.511$

Therefore the value of $log_6{15}$ is 1.511.

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Jorge bought 6 large balloons for a party. This was 25% of all the balloons he bought. How many total balloons did Jorge buy sho

riadik2000 [5.3K]

Answer:

The answer would be 24 balloons in total.

Step-by-step explanation:

Jorge bought 6 large balloons, which was 25% of ALL the balloons he bought.

Meaning that since 6 is 25% of the number 24, he bought 24 balloons in total.

PROOF:

25% * 24 = 6

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3 years ago

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Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

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2 years ago

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