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andreev551 [17]

3 years ago

10

On each graph, one figure is a dilation of the other. Find the scale factor of each dilation and classify it as an enlargement o

r as a reduction.

1 answer:

Vera_Pavlovna [14]3 years ago

6 0

Where are the graphs

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A plane lamina with constant density p(x, y) = p occupies a region in the xy-plane bounded by a simple closed path C. Show that

stira [4]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

8 0

4 years ago

the equation of the line a is given as y = -3/4 x -2. Suppose the line a is parallel to line b and lime t is perpendicular to li

Burka [1]

Answer: The equation for line b is $y = -\frac{3}{4}x + 5$

The equation for line t is $y= \frac{4}{3} x+5$

The attachment shows the original line in green, the parallel line b in red, and the perpendicular line t in blue.

Step-by-step explanation: Use slope intercept form, y = mx + b

The given coordinate, (0,5) is on the y-axis, so 5 will be the y-intercept of the equation. Keep the slope of the original equation and substitute the new value for "b"

To write the equation for line t

Both lines share the y-intercept, (0,5) so the "b" term will be the same. The slope of a perpendicular line is the reciprocal of the original slope with the opposite sign.

So invert 3/4 to become 4/3 and change the negative sign to positive.

The resulting equation is y = 4x/3 + 5

3 0

3 years ago

2 right triangles have identical angle measures but different side lengths. The first triangle has side lengths of 6, 10, 8 and

QveST [7]

Answer:

dilation is the answer

6 0

3 years ago

the population of the word grows exponentially at a rate of 1.1 per year. find the number of years it takes for the population t

alexgriva [62]

Answer:

8.4 you just add it so i hope it helps

6 0

3 years ago

In the third week of July, a random sample of 40 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. A

suter [353]

Answer:

a. [6.6350,7.3950]

b. ME=0.5150

Step-by-step explanation:

a. Given that n=40, $\bar x=6.88$ and that: $z_{\alpha/2}=z_{0.05}=1.645$

The required 90% confidence interval can be calculated as:

$\bar x\pm(margin \ of \ error)\\\\\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\6.88\pm(1.645\times \frac{1.98}{\sqrt{40}})\\\\=[6.3650,7.3950]$

Hence, the 90% confidence interval for the population mean cash value of this crop is [6.6350,7.3950]

b. The margin of error at 90% confidence interval is calculated as:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\=(1.645\times \frac{1.98}{\sqrt{40}})\\\\=0.5150$

Hence, the margin of error is 0.5150

8 0

4 years ago