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4 years ago

Can someone help me with this ASAP!!! And show the work plsss!

Mathematics

1 answer:

cupoosta [38]4 years ago

5 0

Answer:

as soon as possible

Step-by-step explanation:

and it an 90 degree angle

u have to add it

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Solve three-fourths b minus one-sixth equals one-half for b

Snowcat [4.5K]

3/4b-1/6=1/2×b
1.75b=.3
b=0.2
I believe this is right. But not 100%

7 0

3 years ago

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is

Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = $\frac{dx}{dt}$

ship B is sailing north at 20 km/h = $\frac{dy}{dt}$

here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find $\frac{dz}{dt}$ at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

$z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\$

$z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h$

derivative first equation w . r. to t we get

$2z\frac{dz}{dt} =$ $-2(130-x)\frac{dx}{dt}$ $+2y\frac{dy}{dt}$

$\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]$

$\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }$

$= \frac{1100}{85.44}\\ = 12.87km/h$

8 0

3 years ago

Solve each of the following systems using the linear combination method. Check your answers

Dvinal [7]

3x+5y=11 answer : 3x+5y-11=0

Step by step explanation:

3x+5y=11
Move the constant to the left

3x+5y-11=11-11
Eliminate the opposite

3x+5y=11

5y+8y

8 0

3 years ago

Tony used a photocopier to dilate the design for a monorail track system. The figure below shows the design and its photocopy

Darina [25.2K]

Answer:

12 m

Step-by-step explanation:

Given that the design, ABCD, was dilated to get a photocopy, EFGH, a scale factor or ratio was multiplied by the original lengths of the design to get the new measurement of the photocopy.

Thus, we are given the ratio, CD:GH = 2:3.

This means, any of the corresponding lengths of both figures would be in that same ratio.

Using the ratio of the design to the photocopy, 2:3, we can find the length of side EH of the photocopy.

The corresponding side of EH in the design is AD = 8m. Thus, AD to EH = ⅔

$\frac{AD}{EH} = \frac{2}{3}$