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4 years ago

6^x+8=17^2x-2 can I have help

1 answer:

7 0

$6^{x+8}=17^{2x-2}\ \ \ \ \ \ |both\ sides\ log_6\\\\log_66^{x+8}=log_617^{2x-2}\\\\x+8=(2x-2)log_617\\\\x+8=2xlog_617-2log_617\\\\x-2xlog_617=-2log_617-8\\\\x(1-2log_617)=-2log_617-8\ \ \ \ |divide\ both\ sides\ by\ (1-2log_617)\\\\x=\frac{-2log_617-8}{1-2log_617}\\\\x=\frac{-(log_617^2+8)}{-(log_617^2-1)}\\\\x=\frac{log_6289+8}{log_6289-1}$

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The cost for 16 ounces of cheese is $3.20. What is the cost for 20 ounces of cheese?

9966 [12]

Answer:

It costs $4 dollars for 20 ounces of cheese.

Step-by-step explanation:

First we find out how many dollars for 1 ounce of cheese.

3.20 ÷ 16 = 0.2

Next we multiply 0.2 by 20.

20 * 0.2 = 4

So, the answer is 4 dollars.

Hope this helps! :)

8 0

3 years ago

DUE IN A HOUR!!!!!!!!!!

viva [34]

Answer:

y=1/4x+7/4

Step-by-step explanation:

5 0

3 years ago

1) find the equation of the line parallel to

In-s [12.5K]

Answer:

1) The equation of the line parallel to x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is 2y = 5x -12

Solution:

1) find the equation of the line parallel to x-5y=6 and through (4,-2).

Given, line equation is x – 5y = 6

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}$

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

$y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on line. }$

$y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14$

Hence, the line equation is 5y = x -14

2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)

$\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15$

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}$

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line $\times$ slope of given line = -1

slope of required line = $-1 \times \frac{5}{-2}=\frac{5}{2}$

And it passes through (2, -1)

Now, using point slope form

$\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}$

Hence, the line equation is 2y = 5x -12

4 0

3 years ago

A soap maker made 1/8 pounds of soap. He split the soap into five pieces. How much does each piece weigh

Tems11 [23]

Answer:

1/40 pounds

Step-by-step explanation:

1/8 ÷ 5 = 1/40

7 0

3 years ago

Read 2 more answers

In general, the probability that a blood donor has Type A blood is 0.40.Consider 8 randomly chosen blood donors, what is the pro

postnew [5]

Answer:

The probability that more than half of them have Type A blood in the sample of 8 randomly chosen donors is P(X>4)=0.1738.

Step-by-step explanation:

This can be modeled as a binomial random variable with n=8 and p=0.4.

The probability that k individuals in the sample have Type A blood can be calculated as:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{8}{k} 0.4^{k} 0.6^{8-k}\\\\\\$

Then, we can calculate the probability that more than 8/2=4 have Type A blood as:

$P(X>4)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\\\\\P(x=5) = \dbinom{8}{5} p^{5}(1-p)^{3}=56*0.0102*0.216=0.1239\\\\\\P(x=6) = \dbinom{8}{6} p^{6}(1-p)^{2}=28*0.0041*0.36=0.0413\\\\\\P(x=7) = \dbinom{8}{7} p^{7}(1-p)^{1}=8*0.0016*0.6=0.0079\\\\\\P(x=8) = \dbinom{8}{8} p^{8}(1-p)^{0}=1*0.0007*1=0.0007\\\\\\\\P(X>4)=0.1239+0.0413+0.0079+0.0007=0.1738$

3 0

3 years ago