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3 years ago

If B is the midpoint of AC, and AC = 8x – 20, find BC. 3x - 1 A B C

Mathematics

1 answer:

sammy [17]3 years ago

6 0

Answer:

x = 9

Step-by-step explanation:

Since B is the midpoint, that means segment AB has to be the same length as BC.

Therefore:

$2(3x-1)=8x-20$

$6x-2=8x-20$

$18=2x$

$9=x$

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There are 12 participants running in a 110m hurdles race. How many ways are there to award a gold, silver, and bronze medal?

prohojiy [21]

Answer:

220 ways

Step-by-step explanation:

The order matters because there's going to be a first, second, and third place.

So, 12P3=12!/(12-3)!*3!=12!/9!*3!=12*11*10*9!/9!*3!=12*11*10/6=220

Therefore, 220 ways

8 0

2 years ago

Help please?

Lemur [1.5K]

Answer:

4/5 to the 5th power

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

What is the measure of the missing angle in the figure below?

Aleks [24]

Sum of all interior angles in a triangle = 180
let missing angle = x

x + 144 + 13 = 180
x + 157 = 180
x = 23

answer
23 degrees

7 0

2 years ago

Read 2 more answers

I NEED HELP ON QUESTION B AND C ASAP.

ipn [44]

Answer:

See below

Step-by-step explanation:

Assumed the dimensions of the top and bottom parts are identical.

Since the cylindrical part has total height of 1.8 cm and the hemisphere volume is transferred to bottom part and the cone part is still full, the value of h is the difference of the total height of cylinder and full part of the top section of cylinder:

h = 1.8 cm - 0.3 cm = 1.5 cm

Find the volume of sand in the bottom part. It consists of a hemisphere and a cylinder of 1.5 cm height.

V(cylinder) = πr²h = 3.14*(2.6/2)²*1.5 ≈ 7.96 cm³
V(hemisphere) = 2/3πr³ = 2/3*3.14*(2.6/2)³ ≈ 4.6 cm³

<u>Total sand in the bottom part:</u>

7.96 + 4.6 = 12.56 cm³

<u>Time taken:</u>

12.56 / 0.05 = 251.2 seconds = 4 min 11.2 seconds

8 0

2 years ago

a spherical balloon is deflated at a rate of 256pi/3 cm^3/sec. at what rate is the radius of the balloon changing when the radiu

Hunter-Best [27]

What you need to look for is $\frac { dr }{ dt }$ when r=8.

Now:

$Volume\quad of\quad a\quad sphere:\\ \\ V=\frac { 4 }{ 3 } \pi { r }^{ 3 }$

$\therefore \quad \frac { dV }{ dr } =\frac { 4 }{ 3 } \pi \cdot 3{ r }^{ 2 }=4\pi { r }^{ 2 }\\ \\ \therefore \quad \frac { dr }{ dV } =\frac { 1 }{ \frac { dV }{ dr } } =\frac { 1 }{ 4\pi { r }^{ 2 } }$

$And:\\ \\ \frac { dV }{ dt } =-\frac { 256\pi }{ 3 } \\ \\ Therefore:\\ \\ \frac { dr }{ dt } =\frac { dr }{ dV } \cdot \frac { dV }{ dt }$

$\\ \\ =\frac { 1 }{ 4\pi { r }^{ 2 } } \cdot -\frac { 256\pi }{ 3 } \\ \\ =-\frac { 256 }{ 12 } \cdot \frac { \pi }{ \pi } \cdot \frac { 1 }{ { r }^{ 2 } }$

$\\ \\ =-\frac { 64 }{ 3{ r }^{ 2 } } \\ \\ When\quad r=8,\\ \\ \frac { dr }{ dt } =-\frac { 64 }{ 3\cdot { 8 }^{ 2 } } =-\frac { 1 }{ 3 } \\ \\ Answer:\quad -\frac { 1 }{ 3 } \quad cm/sec$

3 0

3 years ago