All categories

3 years ago

What is the volume of this object?

Mathematics

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

Step-by-step explanation:

(Assuming that each cube is 1 unit), the volume is 5 + 6 + 8 which equals 19

You might be interested in

Need help will mark brainliest

Black_prince [1.1K]

Answer:

Step-by-step explanation:

180-52/2=128/2=64

5 0

3 years ago

Read 2 more answers

Brock exercises every day. The function E = 15m + 5 represents the number of sit-ups he can do in m minutes. After drinking a ne

adoni [48]

Answer:

$S=45m+15$

Step-by-step explanation:

we have

E------>the number of sit-ups

m----> the time in minutes

S----> the number of sit-ups after drinking the sports drink

$E=15m+5$ ----> equation A

$S=3E$ ----> equation B

substitute the equation A in the equation B

$S=3(15m+5)$

$S=45m+15$

7 0

3 years ago

mrs.beluga is driving on a snow covered road with a drag factor of 0.2. she brakes suddenly for a deer. The tires leave a yaw ma

Aneli [31]

First, we are going to find the radius of the yaw mark. To do that we are going to use the formula: $r= \frac{c^2}{8m} + \frac{m}{2}$
where
$c$ is the length of the chord
$m$ is the middle ordinate
We know from our problem that the tires leave a yaw mark with a 52 foot chord and a middle ornate of 6 feet, so $c=52$ and $m=6$ . Lets replace those values in our formula:
$r= \frac{52^2}{8(6)} + \frac{6}{2}$
$r= \frac{2704}{48} +3$
$r= \frac{169}{3} +3$
$r= \frac{178}{3}$

Next, to find the minimum speed, we are going to use the formula: $s= \sqrt{15fr}$
where
$f$ is <span>drag factor
</span> $r$ is the radius
We know form our problem that the drag factor is 0.2, so $f=0.2$ . We also know from our previous calculation that the radius is $\frac{178}{3}$ , so $r= \frac{178}{3}$ . Lets replace those values in our formula:
$s= \sqrt{(15)(0.2)( \frac{178}{3}) }$
$s= \sqrt{178}$
$s=13.34$ mph

We can conclude that Mrs. Beluga's minimum speed before she applied the brakes was 13.34 miles per hour.

8 0

3 years ago

△MAH∼△WCF . What is the value of x?

spayn [35]

Are there any choices that i can help with to solve the work problem

8 0

3 years ago

Read 2 more answers

What is the answer please.... need help

mina [271]

Answer:

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

$f\left(x\right)=x^3-6x^2+3x+10$

As the highest power of the x-variable is 3 with the leading coefficients of 1.

So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

$f\left(x\right)=x^3-6x^2+3x+10$

$0=x^3-6x^2+3x+10$ ∵ $f(x)=0$

$Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

$\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x=-1,\:x=2,\:x=5$

Therefore, the zeros of the function are:

$x=-1,\:x=2,\:x=5$

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.

8 0

3 years ago