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mylen [45]
4 years ago
7

For each of the following, state the equation of a perpendicular line that passes through (0, 0). Then using the slope of the ne

w equation, find x if the point P(x, 4) lies on the new line. a. y=3x-1 b. y=1/4 x+2
Mathematics
1 answer:
Irina-Kira [14]4 years ago
3 0

Answer:

Below in bold letters.

Step-by-step explanation:

a. y = 3x - 1.

The equation of the perpendicular line will have slope -1/ 3.

So we have the equation y - y1 = -1/3(x - x1)  where (x1, y1) is a point on this perpendicular line.

As the line passes through (0,0):

y - 0 = 1/3(x - 0)

y = -1/3x is the required equation.

If  x,4 is on the line then x  is found by substituting for y:

4 = -1/3 x

x = 4 / -1/3

x = 4*-3 = -12.

b. y = 1/4x + 2

By the same method as in part a:

the equation is y = -4x.

x = 4 / -4 = -1.

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The air distance between santa barbara and bakersfield is 85 miles a private pilot made this flight travel at an average rate of
AveGali [126]

Answer:

The flight took 34 minutes to arrive Bakersfield wich is 85 miles far from Santa Barbara, at a speed of 150 \ mi/hr

Step-by-step explanation:

Hi

<u>Formula</u>

time=\frac{distance}{speed}

<u>Knowns</u>

speed=150\ mi/hr and distance=85\ mi

So we can apply formula time=\frac{85\ mi}{150\ mi/hr}=17/30 \ hr, then, \frac{17hr}{30} \times \frac{60min}{1hr}=34min, is the time the flight took in minutes

5 0
3 years ago
Biologists tagged 139 fish in a lake on January 1, On February 1, they returned and collected a random sample
Bogdan [553]

Answer:

417 fish

Step-by-step explanation:

139/(10/30)=139*3=417

4 0
3 years ago
The region r is enclosed by the curves y=4-4x2 and y=0
Misha Larkins [42]

The region r is enclosed by the curves y₁ = 4 - 4x² and y₂ = 0 is 16/3 or 5.333 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The region r is enclosed by the curves y₁ = 4 - 4x² and y₂ = 0

The intersection points will be

       y₁ = y₂

4 - 4x² = 0

        x = ±1

Then the area bounded by the curves will be

\rm Area = \int _{-1}^1 (y_1-  y_2) dx\\\\Area = \int _{-1}^1 (4 - 4x^2) dx\\\\Area = \left [ 4x  - \dfrac{4x^3}{3} \right ]_{-1}^1\\\\Area = 4 \left ( 1 + 1 \right ) - \dfrac{4}{3} \left ( 1^3 - (-1)^3 \right )\\\\Area = 8 - \dfrac{8}{3}\\\\Area = \dfrac{16}{3} = 5.333 \

More about the area bounded by the curve link is given below.

brainly.com/question/24563834

#SPJ4

6 0
2 years ago
Can anyone help me, with at least the first one?
kirza4 [7]
3. r^2 + 2r - 35               4. a^2- 11a + 28
    (r + 7)(r - 5)                    (a - 7)(a - 4)

5. m^2 - 6m - 7                6. m^2 - m - 2
    (m + 1)(m - 7)                   (m - 2)(m + 1)

7. n^2 - 4n + 3                  8. b^2 - 4b - 5
    (n - 3)(n - 1)                       (b - 5)(b + 1)


3 0
3 years ago
1. The scale of a map is 1:1000. What are the actual dimensions of a
kakasveta [241]

Answer:

Step-by-step explanation:

4 cm × 1000 = 4000 cm = 40 meters

3 cm × 1000 = 3000 cm = 30 meters

A 4 cm by 3 cm  section on the map represents 40 m by 30 m.

area = 40 m × 30 m = 1200 m²

6 0
3 years ago
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