All categories

2 years ago

Which three pairs of measurements are possible side lengths for the triangle?

Mathematics

1 answer:

wlad13 [49]2 years ago

4 0

Answer:

Options (1), (5), and (6)

Step-by-step explanation:

See attached image

You might be interested in

What is the factored form of the polynomial 27x2y-43xy2

iogann1982 [59]

Answer:

xy(27x - 43y)

Step-by-step explanation:

7 0

3 years ago

I need help on this asap please. <br> im being timed.<br> and i put in a photo of the problem.

Elden [556K]

$F=ma\\\\a.\\ma=F\qquad|\text{divide both sides by}\ a\neq0\\\\m=\dfrac{F}{a}\\\\b.\\F=-30,\ a=6.\ \text{Substitute}\\\\m=\dfrac{-30}{6}=-5\\\\c.\\ma=F\qquad|\text{divide both sides by}\ m\neq0\\\\a=\dfrac{F}{m}\\\\F=30,\ m=10.\text{Substitute}\\\\a=\dfrac{30}{10}=3$

6 0

2 years ago

Find the volume of the solid bounded by z = 2 - x2 - y2 and z = 1. Express your answer as a decimal rounded to the hundredths pl

butalik [34]

Answer:

The answer is " $(\frac{\pi}{2})$ ".

Step-by-step explanation:

$z = 2 - x^2 - y^2.........(1) \\\\z = 1.............(2)$

Let add equation 1 and 2:

Using formula: $x^2+y^2=1 \\\\$

convert to polar coordinates

$r=2$

$\theta \varepsilon (0=\pi)=z\\\\V=\int^{2\pi}_{\theta=0}\int^{1}_{\pi=0}\int^{z_{2}}_{z_1} r \ dr \d \theta\\\\$

$=\int^{2\pi}_{0}\int^{1}_{0} (Z_2-z_1)r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} 1- (-1-x^2-y^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (+1 \pm r^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (-r^3 + r) \ dr \d \theta\\\\=\int^{2\pi}_{0} (-\frac{r^4}{4}+\frac{r^2}{1})^{1}_{0} \d \theta\\\\=\int^{2\pi}_{0} (\frac{1}{4}) \d \theta\\\\=(\frac{2 \pi}{4}) \\\\=(\frac{\pi}{2}) \\\\$

8 0

2 years ago

What is the value of | 3 |?

pshichka [43]

The answer is 3 since you only move 3 spots from 0

4 0

2 years ago

Ldentify the following sections of the textbook.

Tema [17]

I don’t know what the drop down options are so I can’t answer your question. Sorry but can u pls help answer my question I just posted? That would be very helpful thx!

7 0

3 years ago