Point D(-1, 5) is reflected across the x-axis. Which statements about D' are true?

Mathematics

1 answer:

vodomira [7]2 years ago

5 0

The reflected point is D'(-1,-5)

For any point (x,y) reflected in the x-axis the image is (x,-y)

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A line intersects the points(-22,-14) and (-18,-12). What is the slope-intercept equation for this line?

zaharov [31]

Hello there,

Well we are going to start off with the equation to find the slope based on the given points:

$\frac{y_{2} - y_{1} }{x_{2} -x_{1} }$

Now using the two given points we are going to plug in and solve:

$\frac{(-12)-(-14)}{(-18)-(-22)}$ = $\frac{2}{4}$ = $\frac{1}{2}$

From this you know that $\frac{1}{2}$ is the slope of the equation. However, to find the y-intercept we are going to use y = mx+ b and plug in one of the points to solve:

(-14) = $\frac{1}{2}$ (-22) + b

(-14) = (-11) + b

-3 = b

That means that the y-intercept is at (0, -3). Lastly, we are just going to plug all this into the slope-intercept form:

y = $\frac{1}{2}$ - 3

Hope I helped,

Amna

6 0

2 years ago

PLEASE HELP!<br>Calculate $40^{13} \pmod{85}.$

SVEN [57.7K]

In this type of calculations, we decompose 13 by checking the lowest powers of the base, that is 40. for example we check 40^2, or 40^3 and compare it to 85

Notice

40*40*40=64,000

so we check how many time does 85 fit into 64,000:

64,000/85=752.94

85*753=64,005; 64000-64,005=-5

this means that

$40^{3} =-5\pmod{85}$

thus

$40^{13} =40^{3*4+1}={(40^{3})}^{4}*40=(-5)^{4}*40 \pmod{85}=\\\\625*40\pmod{85}=(7*85+30)*40\pmod{85}=30*40\pmod{85}\\\\=1200\pmod{85}=(14*85+10)\pmod{85}=10\pmod{85}$

Answer: 10 (mod85)

Remark, the set of all solutions is:

{......-75, 10, 95, .....}, that is 85k +10

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3 years ago

What is the value of x?

kykrilka [37]

Answer:

Step-by-step explanation:

triangle adds upto 180 degrees

180 - (53+45) = 98

Then two lines adapt to 180 degrees

therefore, 180 -98 = 82

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2 years ago