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Katyanochek1 [597]
3 years ago
9

Plz work out DUE TOMORROW

Mathematics
2 answers:
stira [4]3 years ago
7 0
6/3 = 2/1

2/1 is literally 2

hope this helps

your answer is 2

natulia [17]3 years ago
6 0
6/3 simplified is 2/1

Also known as just 2.
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Phyllis is going to take a 5-hour-17-minute flight from Orlando, Florida, to Las
Rainbow [258]

Answer:

Step-by-step explanation:

ok...so she leaves Orlando at 11:05 am.....the flight takes 5 hr and 17 minutes..

11:05 am + 5 hrs = 4:05 pm.....4:05 + 17 minutes = 4:22 pm.

so she arrives at 4:22 pm eastern time....(orlando time)...but Las Vegas is pacific time

so if 12 pm eastern = 9 am pacific.....a difference of 3 hrs

so the local time in Las Vegas when her plane lands is :

4:22 pm - 3 hrs = 1:22 pm <===

6 0
3 years ago
Complete the equation of the lines whose slope is 5 and y-intercept is (0,4)<br>y = ?
Dimas [21]

Answer:

y = 5x + 4

Step-by-step explanation:

y - 4 = 5(x - 0)

That is point-slope form. I'm not sure if you want it in slope intercept form, but slope intercept form of the equation is -

y = 5x + 4

7 0
3 years ago
The elevation of city a is 80 feet below sea level, the elevation of city c is 16 feet below sea level. The elevation of city D
tekilochka [14]
I am almost positive that the answer could be 8 feet below sea level
6 0
3 years ago
Two isosceles triangles share the same base. Prove that the medians to this base are collinear. (There are two cases in this pro
Anna71 [15]

Answer:

Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.

To prove: Point A,D,P are collinear.

Proof:

→Case 1.  When vertices A and P are opposite side of Base BC.

In Δ ABD and Δ ACD

AB= AC   [Given]

AD is common.

BD=DC  [median of a triangle divides the side in two equal parts]

Δ ABD ≅Δ ACD [SSS]

∠1=∠2 [CPCT].........................(1)

Similarly, Δ PBD ≅ Δ PCD [By SSS]

 ∠ 3 = ∠4 [CPCT].................(2)

But,  ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]

2 ∠2 + 2∠ 4=360° [using (1) and (2)]

∠2 + ∠ 4=180°

But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.

So, point A, D, P lies on a line.

CASE 2.

When ΔABC and ΔPBC lie on same side of Base BC.

In ΔPBD and ΔPCD

PB=PC[given]

PD is common.

BD =DC [Median of a triangle divides the side in two equal parts]

ΔPBD ≅ ΔPCD  [SSS]

∠PDB=∠PDC [CPCT]

Similarly, By proving ΔADB≅ΔADC we will get,  ∠ADB=∠ADC[CPCT]

As PD and AD are medians to same base BC of ΔPBC and ΔABC.

∴ P,A,D lie on a line i.e they are collinear.




3 0
3 years ago
Draw an area model 1.1x2.4
Alexus [3.1K]
You have to draw a rectangle and label the longer side 2.4 and the shorter side 1.1

5 0
3 years ago
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