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3 years ago

7

A particular computer takes 13 minutes to download a 39 minute TV show. How long will it take the computer to download a 2 hour

movie?

1 answer:

ddd [48]3 years ago

5 0

Answer:

40mins

Step-by-step explanation:

Change 2hrs into minutes first=120mins

if 39mins=13mins

120mins=?

Then you cross multiply

$\frac{120mins}{39mins} \times 13mins \\ = 40mins$

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Someone help me ASAP!!!!!!

Ludmilka [50]

I am terrible at math, so I am not completely sure!
Buuuut......I think the answer is A=21

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3 years ago

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PLEASE HELP! I HAVE TWO QUESTIONS AND WILL GIVE 100PTS AND BRAINLIEST IF CORRECT!

Volgvan

Answer:

1) $283.50

2) 65.21 meters.

Step-by-step explanation:

Part 1)

We know that a spool of ribbon holds 6.75 meters.

A craft club buys 21 spools.

So, the total length of ribbon bought is:

$T=21(6.75)=141.75\text{ meters}$

Each meter costs $2. So, the total cost will be:

$141.75(2)=\$ 283.50$

Part 2)

A spool holds 6.75 meters.

We bought 21 of them, so that total length we have is 141.75 meters.

The club used 76.54 meters.

So, the amount of ribbon that is left is represented by:

$141.75-76.54$

Subtract:

$=65.21\text{ meters}$

So, 65.21 meters of ribbon will be left.

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3 years ago

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Carson can choose one vehicle (car or truck), one color (blue, red, or silver), and one type of transmission (standard or automa

77julia77 [94]

Answer:

12 possible outcomes

Step-by-step explanation:

Choosing 1 vehicle from 2 = 2C1 = 2

Choosing 1 color from 3 = 3C1 = 3

Choosing 1 type from 2 = 2C1 = 2

Number of possible outcomes :

2C1 * 3C1 * 2C1

2 * 3 * 2

= 12 different possible outcomes

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3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

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Answer is needed asap

Oksi-84 [34.3K]

I think it is letter A ,man. yeah that gotta be the answer

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