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2 years ago

O || u find the measure of < 3?

Mathematics

1 answer:

Fudgin [204]2 years ago

8 0

Answer:

vertical angles

∠1 and ∠3 are vertical angles. Their measures are equal, so m∠3 = 90 . ∠1 and ∠2 are supplementary.

Step-by-step explanation:

Remember in every triangle, the sum of the angles adds to 180 degrees. So let us call the smallest angle's measure "A". The largest angle is 3A and the third angle is A+10. So A + 3A + (A + 10) = 180.

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1 year ago

A group of students is timed while sprinting 100 meters. Each student’s speed can be found by dividing 100 m by their time. Is e

vesna_86 [32]

B.Time is a function of distance. is the answer ot the question

A group of students is timed while sprinting 100 meters. Each student’s speed can be found by dividing 100 m by their time. Is each statement true or false?

Speed is a function of time.

Time is a function of distance.

Speed is a function of number of students racing.

Time is a function of speed.

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2 years ago

Please Help Me

astra-53 [7]

P = 2(L + W)
L = 1/3W - 3

P = 2(1/3W - 3 + W) or P = 2(4/3W - 3)

5 0

2 years ago

Someone help plz this is do tonight plzzzz

erik [133]

Answer:

y = 5x - 8

Step-by-step explanation:

Here. we want to find the equation of the line

The general form is;

y = mx + b

where m is the slope and b is the y—intercept

To get the slope, we can pick any two points on the line

These are;

(2,2) and: (3,7)

we have the slope as follows;

m = (y2-y1)/)x2-x1) = (7-2)/3-2) = 5/1 =5

So we have;

y = 5x + b

To get b , use any point on the line and substitute its coordinates

using (3,7)

7 = 5(3) + b

7 = 15 + b

b = 7-15

b = -8

So the equation of the line is (

y = 5x - 8

3 0

2 years ago

Plot the foci of the hyperbola represented by the equation x^2/625 − y^2/3600 =1 100pts

Pavel [41]

The attached image represents the foci of the hyperbola

<h3>How to determine the foci?</h3>

The equation of the hyperbola is given as:

$\frac{x^2}{625} - \frac{y^2}{3600} = 1$

Rewrite as:

$\frac{x^2}{25^2} - \frac{y^2}{60^2} = 1$

A hyperbola is represented as:

$\frac{(x - h)^2}{b^2} - \frac{(y - k)^2}{a^2} = 1$

This means that:

h = 0

k = 0

b = 25

a = 60

Next, calculate c the distance from the center to the focus using:

$c = \sqrt{a^2 -b^2}$

This gives

$c = \sqrt{60^2 -25^2}$

Evaluate

$c = \pm \sqrt{2975}$

This means that:

Foci = (0, -√2975) and (0, √2975)

See attachment for the hyperbola and the foci