All categories

3 years ago

Which line is parallel to the line y=1/5x-1 and passes through the point (-10,-3)

1 answer:

4 0

The slope of the line parallel to another line is the same as that of the other line. Therefore, the slope of the parallel to y= 1/5x-1 is 1/5.

Using the point slope from, y-y1= m(x-x1)
Where y-1 and x1 are points from giving coordinate pair (x,y) that lies one the line, and m is the slope.

Plugging in 1/5 for m, -10 for x1 and -3 for y1:

y-(-3)=1/5(x- -10)
y +3 = 1/5x+2
y=1/5x-1

I hope this is right.

You might be interested in

The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli i

harina [27]

Answer:

$y(t) = y_o e^{\beta t}$

$x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

Step-by-step explanation:

From the question we are told that

$\frac{dy}{y} = -\beta dt$

Now integrating both sides

$ln y = \beta t + c$

Now taking the exponent of both sides

$y(t) = e^{\beta t + c}$

=> $y(t) = e^{\beta t} e^c$

Let $e^c = C$

$y(t) = C e^{\beta t}$

Now from the question we are told that

$y(0) = y_o$

Hence

$y(0) = y_o = Ce^{\beta * 0}$

=> $y_o = C$

$y(t) = y_o e^{\beta t}$

From the question we are told that

$\frac{dx}{dt} = -\alpha xy$

substituting for y

$\frac{dx}{dt} = - \alpha x(y_o e^{-\beta t })$

=> $\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt$

Now integrating both sides

$lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c$

Now taking the exponent of both sides

$x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}$

=> $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c$

Let $e^c = A$

=> $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now from the question we are told that

$x(0) = x_o$

$x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

=> $x_o = K e^{\frac {\alpha y_o }{\beta } }$

divide both side by $(K * x_o)$

=> $K = x_o e^{\frac {\alpha y_o }{\beta } }$

$x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

=> $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=> $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

Generally as t tends to infinity , $e^{- \beta t}$ tends to zero

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

5 0

3 years ago

Need help on this question

ololo11 [35]

45 degree between Oak street and Elm Street.

4 0

3 years ago

HELPP!!! 30 POINTS!!! WILL MARK BRAINLYIST!! Step 1: Written Response (30 points) Using complete sentences, compare the fat cont

yan [13]

Answer:

median a= 65 b= 75 c=68

This shows the median of each fat content of each restaurant showing that restaurant c is closer to that of 'restaurant a ' than a is to 'restaurant b' and near to 1/7 lower than that of restaurant b which shows they are all highly skewed with each other upon their graph.

The measures of spread start with 55 to 72 for 'restaurant a' where the upper and lower quartiles are 70-60 showing a distribution 10-17 into the whisker plot.

Compared to restaurant b which shows a measure of spread start with 65 to 90 for 'restaurant b' where the upper and lower quartiles are 85 - 68 into the whisker plot and show a spread of 3 in the left exterior and 5 spread in the right exterior to the upper quartile. So in comparison to restaurant a that restaurant b has a greater spread in the interquartile range = 15 where restaurant a =10.

For the outer quartile the comparison to restaurant a is twice as small meaning restaurant a is greater by over double and shown as 53 <a < 60 which when showing both outer quartiles we can use the amounts being 7 and 70<a<72 = 2

and show closed dots on equality line number line separately showing a<-7 and a>2 for each outer exterior quartile. For restaurant b this shows opposite similar equalities 65<b< 68 = b>-3 and 82<b<89 = b>7

So the differences are smaller for b for left side by 4 and larger for b by 5 up on the right side.

We compare both to restaurant c and find c = 60<c<61 = c>-1 and 70<c<71 = c>1 so the differences are that the outer quartile for c through the other quartiles = -2> c < -6 . This shows how smaller c is compared to a and b outer quartiles). We can also prove that while the outer quartiles are much more smaller for c than a and b we cna prove that c actually has an inner quartile more similar to c and closer distribution of b as the median is more closer for a and c where b has a greater output for median as restaurant b has the higher fat content and greater distribution within the inner quartiles over all.

Summary findings Restaurant c outer quartile is moderately skewed as they show -1 -0.5 on the left side and 1-0.5 on the right side. Restaurant a has a closer median inner quartile to c and closer distribution of inner distribution of c. It's output outer quartile distribiution is a distribution that is higher than b but a smaller inner quartile compared to b, when this happens then the distribution spread shows less range and fewer products to account for.

So i think restaurant b has the higher fat content to its menu.

Where restaurant a must be the healthiest as it holds the lowest range and larger gap is such range for healthier food.ie) when compared to the others.

Meanings of what we are asked.

In a box and whisker plot: the ends of the box are the upper and lower quartiles, so the box spans the interquartile range. the median is marked by a vertical line inside the box. the whiskers are the two lines outside the box that extend to the highest and lowest observations.

Skewness refers to distortion or asymmetry in a symmetrical bell curve, or normal distribution, in a set of data. If the curve is shifted to the left or to the right, it is said to be skewed. Skewness can be quantified as a representation of the extent to which a given distribution varies from a normal distribution.

As a general rule of thumb:

If skewness is less than -1 or greater than 1, the distribution is highly skewed.

If skewness is between -1 and -0.5 or between 0.5 and 1, the distribution is moderately skewed.

If skewness is between -0.5 and 0.5, the distribution is approximately symmetric.

4 0

3 years ago

What’s 20 divided by 20 im tired

MaRussiya [10]

Answer:

The answer to 20 ÷ 20 is 1

7 0

3 years ago

Read 2 more answers

a person leaves their house to go to the store. it takes 15 minutes at 12 mph to get to the store. They return home at 31 mph. W

trasher [3.6K]

$\bf \cfrac{\stackrel{speed~forth}{12}+\stackrel{speed~back}{31}}{\stackrel{interval~forth}{1}+\stackrel{interval~back}{1}}\implies \cfrac{43}{2}\implies 21.5mph\textit{ on average}$

5 0

3 years ago