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Sav [38]
3 years ago
11

Help um. Order the set of numbers from least to greatest

Mathematics
2 answers:
Sati [7]3 years ago
5 0

Answer:

 2\frac{3}{4}, 2.71, \sqrt{5}, \frac{5}{2}

Step-by-step explanation:

Just convert them all to a decimal, I'm too lazy to write it all out...

Kipish [7]3 years ago
4 0

5/2 , √5 , 2¾ , 2.71 (bar)

am not really sure but i guess that is what I know.

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How do I find the measure of angle ABE??
lana66690 [7]

You find the measure of angle BEC, by adding angle B and angle C which is 43+90, and then subtract that from 180 and get 180-(43+90). Then you take the measure of angle BEC (which is the same as AEB), take AEB and A which is 90 degrees and add those together. Then you take that and subtract it from 180, which is 180-(AEB+90)

3 0
2 years ago
Prove ΔPAB is isosceles.
Licemer1 [7]

Answer:

See explanation

Step-by-step explanation:

If PX\cong PY, then triangle PXY is isosceles triangle. Angles adjacent to the base XY of an isosceles triangle PXY are congruent, so

\angle 1\cong \angle 2

and

m\angle 1=m\angle 2

Angles 1 and 3 are supplementary, so

m\angle 3=180^{\circ}-m\angle 1

Angles 2 and 4 are supplementary, so

m\angle 4=180^{\circ}-m\angle 2

By substitution property,

m\angle 4=180^{\circ}-m\angle 2=180^{\circ}-m\angle 1=m\angle 3

Hence,

\angle 3\cong \angle 4

Consider triangles APX and BPY. In these triangles:

  • PX\cong PY - given;
  • \angle 5\cong \angle 6 - given;
  • \angle 3\cong \angle 4 - proven,

so \triangle APX\cong \triangle BPY by ASA postulate.

Congruent triangles have congruent corresponding sides, then

AP\cong BP

Therefore, triangle APB is isosceles triangle (by definition).

6 0
3 years ago
At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce
Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then \frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }.  By the Mean Value Theorem, there is a number c such that 0 < c with v'(c)=60 \:{\frac{mi}{h^2}}. Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 60 \:{\frac{mi}{h^2}}.

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

  1. f is continuous on the closed interval [a, b].
  2. f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

f'(c)=\frac{f(b)-f(a)}{b-a}

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then v(0 \:h) = 30 \:{\frac{mi}{h} } and v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} } (note that 20 minutes is 20/60=1/3 of an hour), so the average rate of change of v on the interval [0 \:h, \frac{1}{3} \:h] is

\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in (0 \:h, \frac{1}{3} \:h) at which v'(c)=60 \:{\frac{mi}{h^2}}.

c is a time time between 2:00 and 2:20 at which the acceleration is 60 \:{\frac{mi}{h^2}}.

4 0
3 years ago
Camden is collecting bugs for science class the first day sister helps him he finds 35 bugs after day two he has 52 bugs on day
trapecia [35]
I believe it is 204 hopefully this helps
5 0
3 years ago
Read 2 more answers
Solve the following using substitution method
Wewaii [24]

Solution :

2a + 2b = 7    ...1)

4a + 3b = 12   ...2)

In equation, 1)

a = (7 - 2b)/2   ...3)

Putting value of a in equation 2) we get :

4 × (7 - 2b)/2  + 3b = 12

2( 7 - 2b ) + 3b = 12

14 - 4b + 3b = 12

b = 2

Putting value of b in 3) we get :

a = ( 7 - 2×2)/2

a = 3/2 = 1.5

Now,

2x - 3y = 16   ...5)

x + 2y = -6    ...6)

x = -6 - 2y

Putting above value of x in eq 5) , we get :

2( -6 - 2y ) - 3y = 16

-12 - 4y - 3y = 16

7y = -28

y = -4

x = -6 - ( 2× -4 )

x = 2

Hence, this is the required solution.

6 0
2 years ago
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