All categories

2 years ago

Isosceles triangle with one angle of 100. What would most likely be the other angles?

2 answers:

4 0

Isoceles triangles have one pair of congruent base angles opposite their congruent sides. Also, the angles in a triangle add up to 180°. Therefore, we can set up this equation:

100 + x + x = 180

(we know it's not the 100° angle that has a congruent match because that would cause the total angle measure to go above 180°)

100 + 2x = 180
Subtract 100 from each side.
2x = 80
Divide by 2.
x = 40°

Both of the other angles would be 40°.

Sergeu [11.5K]2 years ago

3 0

There's no "most likely" about it.

==> Every isosceles triangle has two equal angles.
==> In this one, the 100-degrees angle is either one of the two equal angles,
or else it isn't.
==> If the 100 is one of the two equal angles, then the other one is also 100,
and the sum of the angles in this triangle ... even without the third one ...
is already more than 180 degrees.
==> So the 100 CAN'T be one of the equal angles. It must be the third angle.

-- Now there are 80 degrees left over to split between the other two angles,
which ARE equal.

-- The other two angles, which are not 100 degrees, MUST be 40 degrees each.

Not probably, not maybe, not could-be, and not "most likely". They must be.

You might be interested in

You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in

emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

$j(t)=95t-9t^2$

$j'(t)=95-18t$

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

-18t=-95

$t=\frac{95}{18}$

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

$j(5.28)=95(5.28)-9(5.28)^{2}$

$j(5.28)=250.69ft$

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

$j(t)=95t-9t^{2}$

$j'(t)=95-18t$

j"(t)=-18

and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

5 0

2 years ago

When dividing with polynomials, the goal is to determine how many times the dividend divides evenly into the?

kkurt [141]

Whether dividing constant terms or polynomials, we always have definitive terms when it comes to division. Suppose we say, 10x divided by 2. The dividend is the 10x and the divisor is the 2. In other words, the dividend is the number to be divided by the divisor, to obtain the answer called the quotient.

When dividing polynomials, your main goal is to be able to divide the dividend evenly into the divisor. For example, we divide x²+2x+1 by x+1. The first thing you're going to focus is, what term will completely divide the first term of the polynomial? That would be x. Why? Because when you multiply x with x+1, the product is x²+x. When you subtract this from the polynomial, the x² will cancel out. All you have to do is subtract x from 2x, yielding x. Then, you carry down the last term of the equation: +1. You do the steps again. The term that will completely divide x+1 by x+1 is 1. When you subtract the two, you will come up with zero. That means there is no remainder. The polynomial is divisible by the divisor.
x + 1
------------------------------------
x+1| x²+2x+1
- x²+x
----------------------
x +1
- x +
------------
0

5 0

2 years ago

The Benton Youth Soccer Team has 20 players on the team, including reserves. Of these, three are goalies. Today, the team is hav

inna [77]

Answer:

Step-by-step explanation:

Subtract one player from the total number, because one has to block.

Multiply this by the number of goalies there are.

19*3=57

There will be 57 kicks.

3 0

1 year ago

A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0

2 years ago

The measure of ∠1 = 43 86 274

german

Answer:

$m$

Step-by-step explanation:

we know that

The inscribed angle measures half that of the arc comprising

$m$

8 0

2 years ago

Read 2 more answers