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3 years ago

Sam's car can travel 320 miles on 16 gallons of gas. Find the unit rate.

Mathematics

2 answers:

Snowcat [4.5K]3 years ago

8 0

Answer:I believe the answer is 20.

Step-by-step explanation:

Softa [21]3 years ago

5 0

The answer is 20 because 320/16 is 20

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Can someone please help me ASAP

deff fn [24]

Answer:

QR = 17 cm

Step-by-step explanation:

Δ RST is a 5- 12- 13 triangle with hypotenuse RT = 13 cm , then

TS = 5 cm and PT = 2 × 5 = 10 cm

So PS = 10 + 5 = 15 cm

PS is parallel to the vertical line from vertex Q and intersects the horizontal line projected from SR of length 20 - 12 = 8 cm

Using the right triangle formed calculate QR using Pythagoras' identity

QR² = 15² + 8² = 225 + 64 = 289 ( take square root of both sides )

QR = $\sqrt{289}$ = 17

3 0

3 years ago

Which point is located at (2,7)?

Anastasy [175]

Point S is located at 2,7.

8 0

3 years ago

Read 2 more answers

Answer choices:

Mamont248 [21]

Answer:

(x+8,y-6)

Step-by-step explanation:

it's the number of values that x and y differ

7 0

3 years ago

Pls help, i forgot what 6 - 4 is pls help ;(

Viefleur [7K]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the square root of -2i?

Studentka2010 [4]

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of $z=0-2i$ . First, convert from Cartesian to polar form:

$r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2$

$\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}$

$z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$

Next, use the formula $\displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr]$ where $\displaystyle k=0,1,2,...\:,n-1$ to find the square roots:

When k=1

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]$

$\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i$

When k=0

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i$

Thus, the square roots of -2i are 1-i and -1+i

4 0

2 years ago