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3 years ago

9 miles 10 miles 19 miles 22 miles ??

Mathematics

1 answer:

Softa [21]3 years ago

7 0

Joan is 5 miles away from Gary.
Gary is 8 miles away from Kash.

Let the distance between Joan and Kash be $x$ . Then you can use Pythagorean theorem to find it:
$x=\sqrt{5^2+8^2}=\sqrt{25+64}=\sqrt{89}\approx 9.434$

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What is the area of trapezoid ABCD ? Enter your answer as a decimal or whole number in the box. Do not round at any steps. units

lidiya [134]

Answer:

37.5 sq. units.

Step-by-step explanation:

The coordinate of vertices of trapezoid ABCD are given,

From the diagram it is clear the area of ABCD = $\frac{1}{2}\times AB \times (AD + BC)$ {Since, AD ║ BC and AB is the perpendicular distance between the parallel lines }

Now, AB = $\sqrt{(- 3 - 1)^{2}+ (2 - 5)^{2} } = 5$ units

AD = $\sqrt{(- 3 - 0)^{2}+ (2 - (-2))^{2} } = 5$ units

Again, BC = $\sqrt{(1-7)^{2}+ (5 - (-3))^{2} } = 10$ units

Therefore, the area of trapezoid ABCD = $\frac{1}{2} (10 + 5) \times 5 = 37.5$ sq. units. (Answer)

We know, that the distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is

$\sqrt{(x_{1} - y_{1} )^{2} + (x_{2} - y_{2} )^{2} }$ .

7 0

3 years ago

Find the MODE for the data set: 8, 4, 6, 6, 8, 2, 6,5

gogolik [260]

Answer: 6

Step-by-step explanation: Mode means most reoccurring number. And so it's 6.

4 0

3 years ago

Read 2 more answers

What is the value of x?

rewona [7]

Answer:

x = 80

Step-by-step explanation:

The sum of the angles of a triangle is 180 degrees

70+30+x = 180

100+x = 180

Subtract 100 from each side

100+x-100 = 180-100

x = 80

6 0

2 years ago

Two-thirds of a number is negative six. Find the number.

Sav [38]

-9 would be your answer

5 0

3 years ago

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Hello, can somebody help me out with this problem? <br> (x)/(x-2)+(x-1)/(x+1)=1

IgorC [24]

$\dfrac x{x-2}+\dfrac{x-1}{x+1}=1$

$\dfrac x{x-2}\cdot\dfrac{x+1}{x+1}+\dfrac{x-1}{x+1}\cdot\dfrac{x-2}{x-2}=1$

So long as $x\neq-1$ and $x\neq2$ , we can carry out the manipulation above. Then

$\dfrac{x(x+1)}{(x-2)(x+1)}+\dfrac{(x-1)(x-2)}{(x+1)(x-2)}=1$

$\dfrac{x(x+1)+(x-1)(x-2)}{(x-2)(x+1)}=1$

$\dfrac{x^2+x+x^2-3x+2}{x^2-x-2}=1$

$\dfrac{2x^2-2x+2}{x^2-x-2}=1$

$\dfrac{2x^2-2x+2}{x^2-x-2}\cdot(x^2-x-2)=1\cdot(x^2-x-2)$

$2x^2-2x+2=x^2-x-2$

$x^2-x+4=0$

We can complete the square to solve:

$x^2-x+\dfrac14+\dfrac{15}4=0$

$\left(x-\dfrac12\right)^2=-\dfrac{15}4$

However, $y^2\ge0$ for all (real) values of $y$ , which means there is no real solution to this equation.

If you're solving over the complex numbers, we can take the square root of both sides to get

$x-\dfrac12=\pm i\dfrac{\sqrt{15}}2$

$\implies x=\dfrac{1\pm i\sqrt{15}}2$

3 0

3 years ago