Ask question

Ask question

All categories

2 years ago

9

Steven has a rectangular rug with a perimeter of 16 feet. The width of the rug is 5 feet.What is the length of the rug

1 answer:

Ipatiy [6.2K]2 years ago

5 0

The Length would be 3

5+5= 10
16-10 =6 / 2 =3

You might be interested in

Ken jumps 1 1/5 metres steve jumps 1.5 metres steve jumps further than ken how much further does steve jump than Ken?

monitta

Answer:0.3 m

Step-by-step explanation:

Given

Ken can Jump $1\ \frac{1}{5}\ m$

Steve can Jump $1.5\ m$

Converting mixed fraction into a fraction

$\Rightarrow 1\ \dfrac{1}{5}=\dfrac{1\times 5+1}{5}\\\\\Rightarrow \dfrac{6}{5}\ m=1.2\ m$

The difference between their Jumps is

$\Rightarrow 1.5-1.2=0.3\ m$

Steve Jumps 0.3 m more than ken

5 0

2 years ago

Help pls l need some help

blagie [28]

The correct answer is C, 6/7

7 0

2 years ago

Please help ASAP I need to do this

Bumek [7]

Answer is 5 sorry if I’m wrong

4 0

2 years ago

A soft drink machine outputs a mean of 27 ounces per cup. The machine's output is normally distributed with a standard deviation

Rzqust [24]

Answer: 0.6065

Step-by-step explanation:

Given : The machine's output is normally distributed with

$\mu=27\text{ ounces per cup}$

$\sigma=3\text{ ounces per cup}$

Let x be the random variable that represents the output of machine .

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x= 21 ounces

$z=\dfrac{21-27}{3}\approx-2$

For x= 28 ounces

$z=\dfrac{28-27}{3}\approx0.33$

Using the standard normal distribution table , we have

The p-value : $P(21$

$P(z$

Hence, the probability of filling a cup between 21 and 28 ounces= 0.6065

8 0

3 years ago

Given sin theta = 1/3 and 0 theta pi/2 find tan 2 theta

Kazeer [188]

Answer:

$\tan2\theta=\pm\dfrac{4\sqrt{2}}{7}$

Step-by-step explanation:

Given: $\sin\theta = \dfrac{1}{3}$

$0\leq \theta \leq \dfrac{\pi}{2}$

Theta lie in first quadrant.

Multiply both sides by 2

$0\leq 2\theta \leq \pi$

2theta lie in I and II quadrant.

$\tan2\theta$ is positive in I and negative in II quadrant.

$\sin\theta=\dfrac{1}{3}$

$\cos\theta=\dfrac{2\sqrt{2}}{3}$

$\sin2\theta=2\sin\theta\cos\theta$

$\sin2\theta=2\cdot \dfrac{1}{3}\cdot \dfrac{2\sqrt{2}}{3}=\dfrac{4\sqrt{2}}{9}$

$\cos2\theta=\pm \sqrt{1-\sin^22\theta}=\pm\dfrac{7}{9}$

$\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}$

$\tan2\theta=\pm\dfrac{4\sqrt{2}}{7}$

Hence, The value of $\tan2\theta=\pm\dfrac{4\sqrt{2}}{7}$

5 0

3 years ago

Read 2 more answers