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3 years ago

Solve for x. Thank you.

Mathematics

1 answer:

Bas_tet [7]3 years ago

4 0

Answer:

D) 50/3

Step-by-step explanation:

6/10 = 10/(6+x-6)

6/10 = 10/x

6x = 100

x = 100/6

x = 50/3

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If david run 3 times he covers a distance of __ yards

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But how many feet is he running each time?

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3 years ago

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Which equation has a slope of -2 and passes through the point (1,-6)?

Naily [24]

Answer:

y=-2x-4

Step-by-step explanation:

Given that,

Slope, m = -2

It passes through the point (1,-6).

We need to find the equation of line that passes throgh the given point. It can be calculated as:

$y-y_1=m(x-x_1)$

Put y₁ = -6, x₁ = 1 and m = -2

$y-(-6)=(-2)(x-1)\\\\y+6=-2x+2\\\\y+2x+6-2=0\\\\y=-2x-4$

Hence, the equation of line is y=-2x-4.

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2 years ago

Solve the system: 5x-7y=-41 -3x-5y=-3

-Dominant- [34]

Answer:

x = -4 , y = 3

Step-by-step explanation:

5x - 7y = -41 ... (i)

-3x - 5y = -3 ... (ii)

Multiplying (i) by -3 and (ii) by 5 ;

-15x + 21y = 123 ... (i)

-15x - 25y = -15 ... (ii)

Subtracting (i) by (ii) ;

0 + 46y = 138

46y = 138

y = 138 ÷ 46 = 3

Returning to equation (ii) ;

-3x - 5(3) = -3

-3x = -3 + 15

-3x = 12

x = -4

8 0

3 years ago

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I need to find the surface area and volume of all three figures. If you could provide what equations you used to I will be grate

Margarita [4]

Answer:

Part 1) Sphere The surface area is equal to $SA=196\pi\ m^{2}$ and the volume is equal to $V=\frac{1,372}{3}\pi\ m^{3}$

Part 2) Cone The surface area is equal to $SA=(16+4\sqrt{65})\pi\ units^{2}$ and the volume is equal to $V=\frac{112}{3}\pi\ units^{3}$

Part 3) Triangular Prism The surface area is equal to $SA=51.57\ mm^{2}$ and the volume is equal to $V=17.388\ mm^{3}$

Step-by-step explanation:

Part 1) The figure is a sphere

a) Find the surface area

The surface area of the sphere is equal to

$SA=4\pi r^{2}$

we have

$r=14/2=7\ m$ ----> the radius is half the diameter

substitute

$SA=4\pi (7)^{2}$

$SA=196\pi\ m^{2}$

b) Find the volume

The volume of the sphere is equal to

$V=\frac{4}{3}\pi r^{3}$

we have

$r=14/2=7\ m$ ----> the radius is half the diameter

substitute

$V=\frac{4}{3}\pi (7)^{3}$

$V=\frac{1,372}{3}\pi\ m^{3}$

Part 2) The figure is a cone

a) Find the surface area

The surface area of a cone is equal to

$SA=\pi r^{2} +\pi rl$

we have

$r=4\ units$

$h=7\ units$

Applying Pythagoras Theorem find the value of l (slant height)

$l^{2}=r^{2} +h^{2}$

substitute the values

$l^{2}=4^{2} +7^{2}$

$l^{2}=65$

$l=\sqrt{65}\ units$

$SA=\pi (4)^{2} +\pi (4)(\sqrt{65})$

$SA=16\pi +4\sqrt{65}\pi$

$SA=(16+4\sqrt{65})\pi\ units^{2}$

b) Find the volume

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2}h$

we have

$r=4\ units$

$h=7\ units$

substitute

$V=\frac{1}{3}\pi (4)^{2}(7)$

$V=\frac{112}{3}\pi\ units^{3}$

Part 3) The figure is a triangular prism

a) The surface area of the triangular prism is equal to

$SA=2B+PL$

where

B is the area of the triangular base

P is the perimeter of the triangular base

L is the length of the prism

Find the area of the base B

$B=\frac{1}{2} (2.7)(2.3)=3.105\ mm^{2}$

Find the perimeter of the base P

$P=2.7*3=8.1\ mm$

we have

$L=5.6\ mm$

substitute the values

$SA=2(3.105)+(8.1)(5.6)=51.57\ mm^{2}$

b) Find the volume

The volume of the triangular prism is equal to

$V=BL$

where

B is the area of the triangular base

L is the length of the prism

we have

$B=3.105\ mm^{2}$

$L=5.6\ mm$

substitute

$V=(3.105)(5.6)=17.388\ mm^{3}$

5 0

3 years ago

Evaluate 1+5.3 2 please answer quickly

Schach [20]

Answer:

1+5.3=6.3

Step-by-step explanation:

not sure what your asking for with the 2

explain what your looking for with the 2 and maybe we can help you further

5 0

3 years ago

Read 2 more answers