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2 years ago

Which point on the x-axis lies on the line that passes through point P and is perpendicular to line MN?

Mathematics

2 answers:

spin [16.1K]2 years ago

5 0

Answer:

( 1, 0)

Step-by-step explanation:

you know that the slope of the MN line is 1/4 and a slope perpendicular to that is -4/1 ( negative and reciprocal ). then you can either draw a line from point p with the slope -4/1 and fine the point ( 1, 0 ) or you can do it the harder way which is plug the numbers into y=mx+b with 0 being b because you know the x-axis answer will have a y value of 0. (-4=-4x+0) and x is 1.

marishachu [46]2 years ago

3 0

Answer:

Step-by-step explanation:

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If the quantity 4 times x times y cubed plus 8 times x squared times y to the fifth power all over 2 times x times y squared is

MA_775_DIABLO [31]

$\dfrac{4xy^3 + 8x^2y^5 }{2xy^2}$

$= \dfrac{4xy^3(1+2xy^2)}{2xy}$

$= 2y^2(1 + 2xy^2)$

$= 2y^2 + 4xy^4$

Answer: a = 1, b 2, c = 1, d = 4

8 0

2 years ago

A researcher in environmental science is conducting a study to investigate the impact of a particular herbicide on fish. He has

Vedmedyk [2.9K]

Answer:

The herbicide has no impact on fish

Step-by-step explanation:

Experimental research includes to divide the concerned group 'fishes' here into 'treatment' & 'control' group, to study impact of a treatment. Hypothesis test the statistical significance of propositions. Null hypothesis signifies no impact, alternate hypothesis signifies impact.

Let averages enzyme level of treatment & control be Xt & Xc

Null Hypothesis [H0] : Xt = Xc, or Xt - Xc = 0
Alternate Hypothesis [H1] : Xt ≠ Xc, or Xt - Xc ≠ 0

If hypothesis test suggests that there is no significant difference between treatment & control group enzyme level, so we reject H0 & accept H1. It implies that the herbicide has no significant impact on fishes enzymes.

8 0

1 year ago

Farmer McHenry is going to sell his tomatoes at the farmer’s market. It costs Farmer McHenry $2.50 in gas, $1 for the stand, and

enyata [817]

Answer:

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Step-by-step explanation:

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3 0

1 year ago

Read 2 more answers

A gardener wants to replant a circular bare spot in a yard with grass. The bare spot has a radius of 8 inches. What is the area

ch4aika [34]

The area of the bare spot the gardener needs to replant is 201 square feet

Step-by-step explanation:

We have to find the area of the circular bare spot

Given

Radius of bare spot = r = 8 inches

The formula for area of a circle is given by:

$Area=A = \pi r^2$

putting the value

$A = 3.14 * (8)^2\\= 3.14 * 64\\=200.96$

Rounding off to nearest foot

201 ft^2

Hence,

The area of the bare spot the gardener needs to replant is 201 square feet

Keywords: Area, radius

Learn more about area of circle at:

#LearnwithBrainly

7 0

2 years ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

2 years ago