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3 years ago

If a cylindrical vessel is 12 cm long and its surface area is 540 cm2, set up an

Mathematics

1 answer:

velikii [3]3 years ago

7 0

Answer:

Option (1)

Step-by-step explanation:

Surface area of a cylindrical vessel is given by the expression,

Surface area = $2\pi r^{2}+2\pi rh$

= $2\pi r(r+h)$

Here, r = Radius of the vessel

h = Length of the vessel

By substituting the values of 'h' and surface area in the expression,

2πr(12 + h) = 540

Therefore, Option (1) will be the correct option.

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Quad DEFG with mZD= (12x - 4),

nadezda [96]

Answer: x = 7°

Explanation:

We know that sum of angles of a quadilateral is = 360°

Therefore ATQ

⇒ (12x-4)+(18x+4)+(15x+10)+(5x) = 360

⇒ (12x + 18x + 15x + 5x) + (-4+4+10) = 360

⇒ 50x + 10 = 360

⇒ 50x = 360-10

⇒ 50x = 350

⇒ x = 350/50

⇒ x = 7

Therefore value of x is 7°

Must click thanks and mark brainliest

5 0

4 years ago

Need this answered ASAP

Fudgin [204]

i think is 3rd one, cuz in the app photomath it said (-1, -9)

so this might be close

5 0

3 years ago

Write an equation that represents a vertical stretch by a factor of 3 and a reflection in the x-axis of the graph of g(x)=|x|.

Dmitry [639]

Answer:

(x,y) ---> (3x,3y) 3 units reflected on the x-axis

3 0

3 years ago

HELPPPP!!!!!!!! 100 points!!!! I need help on 37-39 thanks!

____ [38]

Answer:

Part 37)

a) Point-slope form $y-5=(1)(x-3)$

b) Slope intercept form $y=x+2$

c) Standard form $x-y=-2$

Part 38)

a) Point-slope form $y-2=0.25(x+4)$

b) Slope intercept form $y=0.25x+3$

c) Standard form $x-4y=-12$

Part 39)

a) Point-slope form $y-3=2(x-1)$

b) Slope intercept form $y=2x+1$

c) Standard form $2x-y=-1$

Step-by-step explanation:

Part 37) Line passing through point (3,5) and slope of 1

Part a) Equation of the line in point slope form

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute the given values

$y-5=(1)(x-3)$

$y-5=(x-3)$ -----> equation of the line in point slope form

Part b) Equation of the line into slope intercept form

The equation of the line into slope intercept form is equal to

$y=mx+b$

where

m is the slope

b is the y-intercept

Substitute the given values in the equation and solve for b

$5=(1)(3)+b$

$5=3+b$

$b=5-3$

$b=2$

substitute

$y=x+2$ -----> equation of the line in slope intercept form

Part c) Equation of the line in standard form

The equation of the line in standard form is equal to

$Ax+By=C$

where

A is a positive integer

B and C are integers

we have

$y=x+2$

Convert to standard form

Subtract y both sides

$y-y=x+2-y$

$0=x+2-y$

Subtract 2 both sides

$-2=x+2-y-2$

$-2=x-y$

Rewrite

$x-y=-2$ ----> equation in standard form

Part 38) Line passing through points (-4,2) and (0,3)

Part a) Equation of the line in point slope form

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

Find the slope m

$m=(3-2)/(0+4)=1/4=0.25$

with the slope m=0.25 and point (-4,2)

substitute the given values

$y-2=0.25(x+4)$ -----> equation of the line in point slope form

Part b) Equation of the line into slope intercept form

The equation of the line into slope intercept form is equal to

$y=mx+b$

where

m is the slope

b is the y-intercept

we have

$m=0.25$

$b=3$ -----> the y-intercept is the point (0,3)

Substitute the given values in the equation

$y=0.25x+3$ -----> equation of the line in slope intercept form

Part c) Equation of the line in standard form

The equation of the line in standard form is equal to

$Ax+By=C$

where

A is a positive integer

B and C are integers

we have

$y=0.25x+3$

Convert to standard form

Multiply by 4 both sides to remove the decimal number

$4y=x+12$

Subtract 4y both sides

$4y-4y=x+12-4y$

$0=x+12-4y$

Subtract 12 both sides

$-12=x+12-4y-12$

$-12=x-4y$

Rewrite

$x-4y=-12$ ----> equation in standard form

Part 39) Line passing through points (1,3) and (2,5)

Part a) Equation of the line in point slope form

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

Find the slope m

$m=(5-3)/(2-1)=2/1=2$

with the slope m=2 and point (1,3)

substitute the given values

$y-3=2(x-1)$ -----> equation of the line in point slope form

Part b) Equation of the line into slope intercept form

The equation of the line into slope intercept form is equal to

$y=mx+b$

where

m is the slope

b is the y-intercept

with the slope m=2 and point (1,3)

Substitute the given values and solve for b

$3=(2)(1)+b$

$3=2+b$

$b=3-2=1$

substitute

$y=2x+1$ -----> equation of the line in slope intercept form

Part c) Equation of the line in standard form

The equation of the line in standard form is equal to

$Ax+By=C$

where

A is a positive integer

B and C are integers

we have

$y=2x+1$

Convert to standard form

Subtract y both sides

$y-y=2x+1-y$

$0=2x+1-y$

Subtract 1 both sides

$-1=2x+1-y-1$

$-1=2x-y$

Rewrite

$2x-y=-1$ ----> equation in standard form

8 0

4 years ago

0.423 rational or irrational

alexgriva [62]

It would b rational...hope this helps

7 0

4 years ago

Read 2 more answers