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4 years ago

10

A train leaves new york at 4:00 pm. a second train leaves the same city in the same direction at 6:00 pm. the second train trave

ls 60mph faster than the first. if the second train overtakes the first at 8:00 pm, what is the speed of each of the two trains?

1 answer:

Amiraneli [1.4K]4 years ago

8 0

Let y be train 1 and x train 2

y= 60mph+x

8-4 = 4 hours

((60+x)+x)4=(x+2)

(60+2x)4=(x+2)

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In MON, J, K, and L are midpoints. If JL = 11, LK = 13, and ON = 20, and JL || MN, LK || MO, and JK || ON, what is the length of

In-s [12.5K]

Answer:

The lengths of MN is 22 units, MO is 26 units and JK is 10 units

Step-by-step explanation:

<em>A l</em><em>ine segment</em><em> joining the </em><em>mid-points of two sides</em><em> in a triangle is </em><em>parallel to the third side</em><em> and </em><em>equal to half its length</em>

In Δ MON

∵ J, K, and L are mid-points

∵ JL // MN and LK // MO

∴ L is the mid-point of ON

∴ J is the mid-point of MO

∴ K is the mid-point of MN

∵ J, L are the mid-points of MO and ON

∵ JL is opposite to MN

→ By using the rule above

∴ JL = $\frac{1}{2}$ MN

∵ JL = 11 units

∴ 11 = $\frac{1}{2}$ MN

→ Multiply both sides by 2

∴ 22 = MN

∴ MN = 22 units

∵ K, L are the mid-points of MN and ON

∵ KL is opposite to MO

→ By using the rule above

∴ KL = $\frac{1}{2}$ MO

∵ KL = 13 units

∴ 13 = $\frac{1}{2}$ MO

→ Multiply both sides by 2

∴ 26 = MO

∴ MO = 26 units

∵ J, K are the mid-points of MO and MN

∵ JK is opposite to ON

→ By using the rule above

∴ JK = $\frac{1}{2}$ ON

∵ ON =20 units

∴ JK = $\frac{1}{2}$ (20)

∴ JK = 10 units

∴ The lengths of MN are 22 units, MO is 26 units and JK is 10 units

3 0

3 years ago

Expand the given power by using Pascal’s triangle. (9a - 10b)^6

xxMikexx [17]

Answer:

$531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

Step-by-step explanation:

1 n=0

1 1 n=1

1 2 1 n=2

1 3 3 1 n=3

1 4 6 4 1 n=4

1 5 10 10 5 1 n=5

1 6 15 20 15 6 1 n=6

This is where n is the exponent in

$(x+y)^n$ .

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

Now we want to expand:

$(9a-10b)^6$ or we we can rewrite as $(9a+(-10b))^6$ .

Let's replace $x$ with $(9a)$ and $y$ with $(-10b)$ in the expansion:

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

$((9a)+(-10b))^6$

$=1(9a)^6+6(9a)^5(-10b)+15(9a)^4(-10b)^2+20(9a)^3(-10b)^3+15(9a)^2(-10b)^4+6(9a)(-10b)^5+1(-10b)^6$

Let's simplify a bit:

$=9^6a^6-60(9)^5a^5b+15(-10)^2(9)^4a^4b^2+20(9)^3(-10)^3a^3b^3+15(9)^2(-10)^4a^2b^4+6(9)(-10)^5ab^5+(-10)^6b^6$

$=531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

8 0

4 years ago

Jones Co. pays its employees on a graduated commission scale: 2% on first $30,000; 3% on sales from $30,000 to $60,000; and 5% o

Gnesinka [82]

We need to separate his total earnings out into the different categories:

Category 1 (2%): 30,000
Category 2 (3%): 60,000-30,000 = 30,000
Category 3 (5%): 92,000-60,000 = 32,000

To calculate a percentage we multiply by that percentage divided by 100.
Now we can work out the commissions:
1) 30,000*0.02 = $600
2) 30,000*0.03 = $900
3) 32,000*0.05 = $1600

Adding these together, we get $3100

3 0

3 years ago

Someone pls help me with this and I’ll mark you as brainliest!!

MrRissso [65]

Answer:

2. I am pretty sure that it is right angles are congruent.

4. Angle O and R are congruent, never heard of their angles theorem but I think it might be related to CPCTC.

Step-by-step explanation:

Well a square is always used to represent a right angle so and both angle Q and N have squares and that should mean that they are both 90 degrees. Also while vertical angels are congruent, angles Q and N are not vertical from each other, Reflexive Property is if it is a side and not an angle I think.

5 0

3 years ago

Line XY bisects and is perpendicular to AB and CD. CD = 10 and XY = 24. What is the length of AP?

myrzilka [38]

Answer:

<u>AP = 13</u>

Step-by-step explanation:

See the attached figure.

CD = 10 and XY = 24

Line XY bisects and is perpendicular to AB and CD

∴ XY bisects at Y

∴ CY = 0.5 CD = 0.5 * 10 = 5

And from the geometry of the shape

P is the midpoint of XY

∴ XP = YP = 0.5 XY = 0.5 * 24 = 12

XY is perpendicular to CD

∴ ΔPYC is a right triangle at Y

∴ PC (hypotenuse) = $\sqrt{PY^2+YC^2} =\sqrt{12^2+5^2}=\sqrt{144+25} =\sqrt{169} =13$

From the figure AP = PC = 13

8 0

3 years ago