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2 years ago

What are the coordinates of the point on the directed line segment from (5,5) to

Mathematics

1 answer:

iogann1982 [59]2 years ago

6 0

Answer:

$(x,y) = (6,2)$

Step-by-step explanation:

Given

$(x_1,y_1) = (5,5)$

$(x_2,y_2) = (8,-4)$

$m:n = 1:2$

Required

Determine the coordinate of the partition

This is calculated using:

$(x,y) = (\frac{mx_2 + nx_1}{m + n},\frac{my_2 + ny_1}{m + n})$

Substitute values for m,n,x's and y's

$(x,y) = (\frac{1 * 8 + 2 * 5}{1+2},\frac{1*-4 + 2*5}{1 + 2})$

$(x,y) = (\frac{18}{3},\frac{6}{3})$

$(x,y) = (6,2)$

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X²+y²-6x+14y-1=0 and please show your work so I can learn

Eva8 [605]

Hello there,

I hope you and your family are staying safe and healthy during this winter season.

$x^2 + y^2 -6x+14y-1=0$

We need to use the Quadratic Formula*

$x =\frac{-b+\sqrt{b^2}-4ac }{2a}$ , $\frac{-b-\sqrt{b^2} -4ac }{2a}$

Thus, given the problem:

$a = 1, b=-6, c=y^2+14y-1$

So now we just need to plug them in the Quadratic Formula*

$x=\frac{6+2\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$ , $x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}$

As you can see, it is a mess right now. Therefore, we need to simplify it

$x=\frac{6+2\sqrt{10-y^2-14y} }{2}$ , $x = \frac{6-2\sqrt{10-y^2-14y} }{2}$

Now that's get us to the final solution:

$x=3+\sqrt{10-y^2-14y}$ , $x=3-\sqrt{10-y^2-14y}$

It is my pleasure to help students like you! If you have additional questions, please let me know.

Take care!

~Garebear

3 0

1 year ago

A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co

Luda [366]

Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let $\mu$ = mean weight of coffee in its containers.

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 32 ounces {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, $H_A$ : $\mu$ < 32 ounces {means that the mean weight of coffee in its containers is less than 32 ounces}

The test statistics that would be used here One-sample t-test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = 31.8 ounces

s = sample standard deviation = 0.48 ounces

n = sample of containers = 15

So, the test statistics = $\frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }$ ~ $t_1_4$

= -1.614

The value of t test statistics is -1.614.

Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

8 0

2 years ago

HelpPlease!!! If s(x)= 2-x2 and t(x)=3 which value is equivalent to s•t(-7)

Shalnov [3]

i think the answer is d

8 0

2 years ago

Read 2 more answers

The graph above shows the cost of renting a scooter. If Jessica drove the scooter 2 miles, how much did it cost to rent a scoote

ella [17]

A) 1.00

Just look at where the slope crosses 2 miles and then look left on the y-axis where the price is given

5 0

2 years ago

Given a triangle MTN, prove that

cupoosta [38]

Answer:

$\angle m + \angle t + \angle n = 180$

Step-by-step explanation:

Required

Show that:

$\angle m + \angle t + \angle n = 180^o$

To make the proof easier, I've added a screenshot of the triangle.

We make use of alternate angles to complete the proof.

In the attached triangle, the two angles beside $\angle m$ are alternate to $\angle t$ and $\angle n$

i.e.

$\angle 1 = \angle t$

$\angle 2 = \angle n$

Using angle on a straight line theorem, we have:

$\angle 1 + \angle m + \angle 2 = 180$

Substitute values for (1) and (2)

$\angle t + \angle m + \angle n = 180$

Rewrite as:

$\angle m + \angle t + \angle n = 180$ -- proved

5 0

2 years ago