How do I solve this

What is the greatest common divisor of 385 and 420?

aliina [53]

Answer:

35

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the volume of the cylindrical barrel that has a diameter of 14mm and a height of 4mm. Round your answer to the nearest tent

algol13

Answer: 615.4mm cube

Step-by-step explanation:

The volume of a cylinder is given as πr^2h

where r= radius

h = height

π = 3.14

From the question, diameter is 14mm and height is 4mm.

To get the radius, we divide the diameter by 2 which will give:

= 14mm/2 = 7mm

Volume= πr^2h

= 3.14 × 7^2 × 4

= 3.14 × 49 × 4

= 615.4mm cube

The volume of the cylindrical barrel is 615.4mm cube.

8 0

3 years ago

Need help with this as soon as possible.

harina [27]

Answer:

1, 2

Step-by-step explanation:

The integers, x, that satisfy 1 <= x <= 8 are

1, 2, 3, 4, 5, 6, 7, 8

Now we solve the inequality for x.

-x + 4 >= 2

-x >= -2

x <= 2

Now we see which of the eight integers above make the inequality true.

Only 1 and 2 satisfy the inequality.

8 0

3 years ago

There are 20 red and blue marbles in a bag. s marbles are red how many are blue?

astra-53 [7]

Answer:

Step-by-step explanation:

Step-by-step explanation:

15 r blue marbles

20- 5 = 15

3 0

2 years ago

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Find the equation of the lines parallel and perpendicular to the line 5x+2y=12 through the point (-2,3)

muminat

Answer:

The equation of line parallel to given line and passing through points ( - 2 , 3 ) is 5 x + 2 y + 4 = 0

The equation of line perpendicular to given line and passing through points ( - 2 , 3 ) is 2 x - 5 y + 19 = 0

Step-by-step explanation:

Given equation of line as :

5 x + 2 y = 12

or, 2 y = - 5 x + 12

or , y = $\frac{-5}{2}$ x + $\frac{12}{2}$

Or, y = $\frac{-5}{2}$ x + 6

∵ Standard equation of line is give as

y = m x + c

Where m is the slope of line and c is the y-intercept

Now, comparing given line equation with standard eq

So, The slope of the given line = m = $\frac{-5}{2}$

Again,

The other line if passing through the points (- 2 , 3 ) And is parallel to given line

So, for parallel lines condition , the slope of both lines are equal

Let The slope of other line = M

So, M = m = $\frac{-5}{2}$

∴ The equation of line with slope M and passing through points ( -2 , 3) is

y = M x + c

Now , satisfying the points

So, 3 = $\frac{-5}{2}$ × ( - 2 ) + c

or, 3 = $\frac{10}{2}$ + c

Or, 3 = 5 + c

∴ c = 3 - 5 = - 2

c = - 2

So, The equation of line with slope $\frac{-5}{2}$ and passing through points ( -2 , 3)

y = $\frac{-5}{2}$ x - 2

or, 2 y = - 5 x - 4

I.e 5 x + 2 y + 4 = 0

<u>Similarly</u>

The other line if passing through the points (- 2 , 3 ) And is perpendicular to given line

So, for perpendicular lines condition,the products of slope of both lines = - 1

Let The slope of other line = M'

So, M' × m = - 1

Or, M' × $\frac{-5}{2}$ = - 1

Or, M' = $\frac{-1}{\frac{-5}{2}}$

Or, M' = $\frac{2}{5}$

∴ The equation of line with slope M and passing through points ( -2 , 3) is

y = M' x + c'

Now , satisfying the points

So, 3 = $\frac{2}{5}$ × ( - 2 ) + c'

or, 3 = $\frac{- 4}{5}$ + c'

Or, 3 × 5 = - 4 + 5× c'

∴ 5 c' = 15 + 4

or, 5 c' = 19

Or, c' = $\frac{19}{5}$

So, The equation of line with slope $\frac{2}{5}$ and passing through points ( -2 , 3)

y = $\frac{2}{5}$ x + $\frac{19}{5}$

y = $\frac{2 x + 19}{5}$

Or, 5 y = 2 x + 19

Or, 2 x - 5 y + 19 = 0

Hence The equation of line parallel to given line and passing through points ( - 2 , 3 ) is 5 x + 2 y + 4 = 0

And The equation of line perpendicular to given line and passing through points ( - 2 , 3 ) is 2 x - 5 y + 19 = 0

Answer

4 0

3 years ago