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2 years ago

Consider the points Q(23, 48) and R(7, 62). What is the component form of Vector Q R?

Mathematics

2 answers:

aev [14]2 years ago

8 0

Answer:

The correct option is;

LeftAngleBracket negative 16, 14 RightAngleBracket which is $\left \langle -16, 14 \right \rangle$

Step-by-step explanation:

The given coordinates of the points are;

The starting point, Q(23, 48) and the final point, R(7, 62)

Therefore, we have;

The x component = The x-value of R - The x-value of R = 7 - 23 = -16

The y component = The y-value of R - The y-value of R = 62 - 48 = 14

Therefore, the component form of the vector QR = $\left \langle -16, 14 \right \rangle$

Alika [10]2 years ago

4 0

Answer:

B on Edg.

Step-by-step explanation:

Yay

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Find the area of the figure. (Sides meet at right angles.)

s344n2d4d5 [400]

Answer:

Step-by-step explanation:

A1 = 2(5 x 2) = 20

A2 = 8 x 3 = 24

total = 20 + 24

= 44 ft²

6 0

3 years ago

The length of the hypotenuse of a right angled triangle exceeds the length of the base by 2cm and exceeds twice the length of th

insens350 [35]

Answer: a = 8; b = 15; c = 17

Step-by-step explanation:

c = b + 2 and c = 2a + 1

b = c − 2 and a = $\frac{c-1}{2}$

c² = b² + a²

$c^{2} = (c-2)^{2} + \frac{c-1}{2}\\$

$c^{2}= \frac{4 (c-2)^{2}+(c - 1)^{2} }{4}$

$4c^{2}={4 (c^{2} + 4 -4c)+c ^{2} + 1-2c$

$4c^{2}={4 c^{2} + 16 -16c)+c ^{2} + 1-2c$

$c^{2}-18c+17=0$

$c(c -17)-1(c-17)=0$

$(c -1)(c-17)=0$

c = 1 or c = 17

If c = 1 then b = 1 - 2 = -1, that's not possible

So x = 17

b = 17 - 2 = 15

a = $\frac{17-1}{2} = 8$

Length sides of the triangle are 17 cm, 15 cm and 8 cm.

I saw some of this on a site but I want to make sure its correct so we're gonna apply the pythagorean theory and see if its correct.

$a^{2} + b^{2} = c^{2} \\8^{2} + 15^{2} = 17^{2} \\64 + 225 = 289\\289 = 289\\$

This shows that this is correct.

Hope this helped! Again, some of this was from a site, not from me.

3 0

3 years ago

1. Mr. Samuel had 13 notes in his wallet. They consist of $1. $5 and $10 notes. The total amount of the notes adds up to $93. Ho

bazaltina [42]

Answer:

8 $10 notes

Step-by-step explanation:

I started by working it backwards. $93-$3 equals $90. So we have three of our 13 notes. We need ten more. Well the most $10 notes we can have is 8. if we had 8 $10 notes that would give us $80. $90-$80=10. We can take ten and divide it by five so we have 2 $5 notes.

So in total he would have 8 $10 notes, 2 $5 notes, and $3 one dollar notes which adds up to 13 notes

7 0

2 years ago

Divide 1-k^3 by k-1

UNO [17]

Answer:

k 2 + k + 1

Step By Step:

Cancel the common factor.

Then.

k 2 + k +1 by 1

Gets you k2+k+1

7 0

3 years ago

How to do this question plz answer

ANEK [815]

9514 1404 393

Answer:

more than half are being used

Step-by-step explanation:

Perhaps easiest to understand is the most straightforward: figure the total number of seats and the number of seats being used. Check to see if that is more than half.

The 1 first-class carriage has usage of ...

(3/8)(64 seats) = 24 seats.

The 6 standard carriages have usage of ...

(7/13)(6 · 78 seats) = 252 seats.

The total number of seats used is ...

24 + 252 = 276 . . . . seats used

The total number of seats on the train is ...

1·64 +6·78 = 532 . . . . seats on the train

Half of the seats on the train will be ...

532/2 = 266 seats . . . . half the seats

The 276 used seats are more than half of the total number. More than half the seats are being used.

_____

<em>Alternate solution</em>

Another way to look at this is to compare utilization to 1/2 in each carriage class.

In first class 1/2 -3/8 = 1/8 of 64 seats = 8 seats fewer than 1/2 the seats are being used.

In standard class, 7/13 -1/2 = 14/26 -13/26 = 1/26 of 78 seats = 3 seats more than 1/2 the seats are being used. There are 6 standard class carriages, so 6·3 = 18 more than half the seats in the standard-class carriages are being used.

This number is 10 more than the 8 under-utilized seats in first class, so <em>more than half the train seats are being used</em>.

8 0

2 years ago