All categories

3 years ago

Need answer pleaseeeee

Mathematics

2 answers:

Afina-wow [57]3 years ago

4 0

C 150cm is the volume

geniusboy [140]3 years ago

4 0

Answer:

150cm3

Step-by-step explanation:

multiply all the numbers together and the three stands for the total number of cm were added.

You might be interested in

If a statement is true, then it’s converse must be true as well. True or false

TEA [102]

Answer: true

Explanation: because if a statement is true then it’s convert

5 0

2 years ago

the angle theta is located in quadrant 1 and cos of theta is 3/8. what is the value of sin of theta ?

ra1l [238]

Answer: sin of theta = (√55)/8

Step-by-step explanation:

Cosine is the x-value divided by the hypotenuse. We now know that since cosine is 3/8, that the hypotenuse is 8. In order to find the third side of the triangle (which would be the v-value), we must use Pythagorean theorem. The third side is square root 55. Sine equals the y-value divides by the hypotenuse. We know both of those values, so Sine of theta = square root of 55 divided by 8.

8 0

2 years ago

A triangle has a perimeter of 18 inches. If one side has length 8 inches, find the length of the shortest side if the other leng

aalyn [17]

$a,b,c-\ sides\ of\ triangle\\\\a+b+c=18\ inches\\\\a=8inch\\\\\frac{b}{c}=\frac{2}{3}\\\\3b=2c\ \ --->b=\frac{2c}{3}\\\\8+\frac{2c}{3}+c=18\ |*3\\\\24+2c+3c=54\\\\5c=54-24\\\\5c=30\\\\c=\frac{30}{5}=6\\\\Length\ of\ shorter\ side\ is\ equal\ to\ 6\ inches.$

8 0

2 years ago

What is the length of BC , rounded to the nearest tenth?

Arte-miy333 [17]

Step $1$

In the right triangle ADB

Find the length of the segment AB

Applying the Pythagorean Theorem

$AB^{2} =AD^{2}+BD^{2}$

we have

$AD=5\ units\\BD=12\ units$

substitute the values

$AB^{2}=5^{2}+12^{2}$

$AB^{2}=169$

$AB=13\ units$

Step $2$

In the right triangle ADB

Find the cosine of the angle BAD

we know that

$cos(BAD)=\frac{adjacent\ side }{hypotenuse}=\frac{AD}{AB}=\frac{5}{13}$

Step $3$

In the right triangle ABC

Find the length of the segment AC

we know that

$cos(BAC)=cos (BAD)=\frac{5}{13}$

$cos(BAC)=\frac{adjacent\ side }{hypotenuse}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{13}{AC}$

solve for AC

$AC=(13*13)/5=33.8\ units$

Step $4$

Find the length of the segment DC

we know that

$DC=AC-AD$

we have

$AC=33.8\ units$

$AD=5\ units$

substitute the values

$DC=33.8\ units-5\ units$

$DC=28.8\ units$

Step $5$

Find the length of the segment BC

In the right triangle BDC

Applying the Pythagorean Theorem

$BC^{2} =BD^{2}+DC^{2}$

we have

$BD=12\ units\\DC=28.8\ units$

substitute the values

$BC^{2}=12^{2}+28.8^{2}$

$BC^{2}=973.44$

$BC=31.2\ units$

therefore

the answer is

$BC=31.2\ units$

8 0

3 years ago

Read 2 more answers

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$