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2 years ago

Can you help me with question number 10, please?

Mathematics

2 answers:

Alekssandra [29.7K]2 years ago

8 0

Answer:

(1) 8.3

Step-by-step explanation:

KatRina [158]2 years ago

8 0

Answer:

1) 8.3

Step-by-step explanation:

get rid of the (), which u will get 4x-28=0.3x+0.6+2.11

combine like terms, which u will get 4x-28=0.3x+2.71

move variable to one side and numbers to the other, which you will get

3.7x=30.71

divide both sides by 3.7 which u will get 8.3

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(2x + 3y) + (4x - 3y) = 11+(-5)

Maurinko [17]

Answer:

The correct option should have been $6x=6$ .

Step-by-step explanation:

Given the expression

$\left(2x\:+\:3y\right)\:+\:\left(4x\:-\:3y\right)\:=\:11+\left(-5\right)$

solving the expression

$\left(2x\:+\:3y\right)\:+\:\left(4x\:-\:3y\right)\:=\:11+\left(-5\right)$

Remove parentheses: (a) = a

$2x+3y+4x-3y=11-5$

Group like terms

$3y-3y+2x+4x=11-5$

Add similar elements

$6x=6$

It is clear that not a single given option is $6x=6$ . It means no option is correct. It seems you mistyped the correct options.

The correct option should have been $6x=6$ .

3 0

2 years ago

A density graph for all of the possible times from 50 seconds to 300 seconds

Vika [28.1K]

Answer: D. took quiz :)

Step-by-step explanation:

7 0

3 years ago

find the value of q for which point p(q,2) is the midpoint of the line segment joining the points q(-5,0) and r(-1,4)

Elza [17]

Answer:

step by step solution:

q=(-1+(-5))/2

q=-3

6 0

2 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

How do you do this? I'm confused.

s2008m [1.1K]

So you divide 50 by 2 so there are 25 before him

8 0

3 years ago