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3 years ago

Line AB contains points A(4, 5) and B(9,7). What is the slope of AB?

Mathematics

2 answers:

vlabodo [156]3 years ago

7 0

Answer:

2/5

Step-by-step explanation:

We can find the slope of a line given two points by using

m =(y2-y1)/(x2-x1)

= (7-5)/(9-4)

= 2/5

Vikki [24]3 years ago

3 0

Answer:

2/5

Step-by-step explanation:

We can find the slope of a line given two points by using

m =(y2-y1)/(x2-x1)

= (7-5)/(9-4)

= 2/5

You might be interested in

A researcher wants to construct a 90% confidence interval for the proportion of elementary school students in Seward County who

dezoksy [38]

Answer:

The sample size is $n =68$

Step-by-step explanation:

The population proportion is $\^ p = 0.50$

The margin of error is $E = 0.1$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n = [\frac{1.645 }{0.1} ]^2 * 0.50 (1 - 0.50 )$

=> $n =68$

6 0

2 years ago

Is my answer correct??

Ivanshal [37]

Answer: Last option.

Step-by-step explanation:

For this exercise it is important to remember that a Right triangle is a triangle that has an angle that measures 90 degrees.

By definition, given a Right triangle, if you draw an altitutde from the vertex of the angle that measures 90 degrees (The right angle) to the hypotenuse, the measure of that altitude is the geometric mean between the measures of the two segments of the hypotenuse.

In this case, you have the Right triangle $EFH$ given in the picture attached.

You can notice that $FG$ goes from the vertex of right angle ( $\angle F=90\°$ ) to the hypotenuse $EH$ . Therefore, it divides the hypotenuse into two segments. These are:

$EG$ and $GH$

Therefore, $FG$ is the geometric mean of the segments $EG$ and $GH$ .

6 0

2 years ago

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0

3 years ago

What is the y in 10.3=0.6y

ollegr [7]

Hello, you would divide 10.3 by 0.6 to get 17.166 repeating. So y equals 17.166.

6 0

3 years ago

Consider a prism ABCA'B'C' whose base is an equilateral triangle of side a. The projection of A' onto ABC is the centroid of ABC

Serggg [28]

Answer:

lsu is there to help you with this question

Step-by-step explanation: