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katrin [286]
2 years ago
6

Write this down in standard form equation: slope = -1, (2, 5)

Mathematics
1 answer:
kykrilka [37]2 years ago
6 0

Answer:

x+y=7

Step-by-step explanation:

Hi there!

We are given a slope of -1, and the point (2,5)

We want to write an equation of the line in standard form

  • Standard form is written as ax+by=c, where a, b, and c are free integer coefficients, but a and b cannot be zero, and a cannot be negative

One way to write the equation of the line in standard form is to first write it in slope-intercept form, and then convert to standard form.

  • Slope intercept form is written as y=mx+b, where m is the slope and b is the y intercept

Since we are already given the slope of the line (-1), we can immediately plug it into the equation y=mx+b

y=-1x+b, or y=-x+b

Now we need to solve for b

As the equation should contain the point (2, 5), it should pass through that point; therefore, it is a solution to the equation, and we can use it to help solve for b

substitute 2 as x and 5 as y:

5=-1(2)+b

Multiply

5=-2+b

add 2 to both sides

7=b

Substitute 7 as b:

y=-x+7

Now we have the equation in slope-intercept form, but remember; we want it in standard form.

Standard form has both x and y on one side, so we can add x to both sides to convert to standard form.

x + y = 7

The equation is written in standard form; a and b (the coefficients in front of x and y) are both not zero, and a is not negative. So we are done.

Hope this helps!

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y=20

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Car is driving at 30 kilometers per hour. How far in meters does it travel in 3 seconds?
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25 meters

Step-by-step explanation:

30 km/hr = (30 km/1 hr)*(1000 m/1 km)*(1 hr/60 min)*(1 min/60 sec)

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The speed or rate is r = 8.3333333333 meters per second approximately, and the time is t = 3 seconds, so,

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A helicopter is flying above a town. the local high school is directly to the east of the helicopter at a 20° angle of depressio
guapka [62]
Answer: 4.5 miles

Explanation:

When you draw the situation you find two triangles.

1) Triangle to the east of the helicopter

a) elevation angle from the high school to the helicopter = depression angle from the helicopter to the high school = 20°

b) hypotensue = distance between the high school and the helicopter

c) opposite-leg to angle 20° = heigth of the helicopter

d) adyacent leg to the angle 20° = horizontal distance between the high school and the helicopter = x

2) triangle to the west of the helicopter

a) elevation angle from elementary school to the helicopter = depression angle from helicopter to the elementary school = 62°

b)  distance between the helicopter and the elementary school = hypotenuse

c) opposite-leg to angle 62° = height of the helicopter

d) adyacent-leg to angle 62° = horizontal distance between the elementary school and the helicopter = 5 - x

3) tangent ratios

a) triangle with the helicpoter and the high school

tan 20° = Height / x ⇒ height = x tan 20°

b) triangle with the helicopter and the elementary school

tan 62° = Height / (5 - x) ⇒ height = (5 - x) tan 62°

c) equal the height from both triangles:

x tan 20° = (5 - x) tan 62°

x tan 20° = 5 tan 62° - x tan 62°

x tan 20° + x tan 62° = 5 tan 62°

x  (tan 20° + tan 62°) = 5 tan 62°

⇒ x = 5 tant 62° / ( tan 20° + tan 62°)

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hipotenuse = 4.46 miles

Rounded to the nearest tenth = 4.5 miles

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