Ask question

Ask question

All categories

3 years ago

15

Find the distance between the points (3, 8) and (-1, 9).

1 answer:

BlackZzzverrR [31]3 years ago

4 0

---------------------------------------------------
Formula
---------------------------------------------------
$\text {Distance } = \sqrt{ (X_1-X2)^{2} + (Y_1-Y_2)^{2} }$

---------------------------------------------------
Find the distance
---------------------------------------------------
Given the point (3, 8) and (-1, 9)

$\text {Distance } = \sqrt{ (3-(-1))^{2} + (8 - 9)^{2} }$

$\text {Distance } = \sqrt{(4)^{2} + (-1)^{2} }$

$\text {Distance } = \sqrt{16+ 1 }$

$\text {Distance } = \sqrt{17 }$

$\text {Distance } = 4.12 \ units$

You might be interested in

Help me fill in the blanks pleasee

777dan777 [17]

Answers and Step-by-step explanations:

Let's look at the first row:

cost = _______ + _______ * (monthly payment)

The first blank should be the "down payment" because this is the initial, constant value that is paid. The second blank should be the "number of months" because this way, when multiplied by the actual amount each payment is (monthly payment), then we get the total amount paid during those months.

So, we have:

cost = (down payment) + (number of months) * (monthly payment)

Now for the second row, we just plug in numbers and variables:

335 = 50 + 6 * p

Finally, we can solve for p:

335 = 50 + 6p

6p = 285

p = $47.50

8 0

2 years ago

Read 2 more answers

For the following problem, assume that all given angles are in simplest form, so that if A is in QIV you may assume that 270° &l

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

Given that angle A is in IV quadrant

So A/2 would be in II quadrant.

sin A = -1/3

cos A = $\sqrt{1-sin^2 A} =\frac{2\sqrt{2} }{3}$

(cos A is positive since in IV quadrant)

Using this we can find cos A/2

$cosA = 2cos^2 \frac{A}{2} -1\\Or cos \frac{A}{2} =-\sqrt{\frac{1+cosA}{2} } =-\sqrt{\frac{3+2\sqrt{2} }{6} }$

3 0

2 years ago

Dada la siguiente serie numérica: ..., –7, -4, -1, 2, 5, ... ¿Cuál es la suma del número entero

Ymorist [56]

Answer:

-5

Step-by-step explanation:

el número anterior a -7 es -10 y -10+5 es -5

4 0

3 years ago

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

$y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0$

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

$p(x) = \dfrac{(x+2)}{(x^2-4)^2} \$

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

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

Also;

$q(x) = \dfrac{7}{(x^2-4)^2}$

$q(x) = \dfrac{7}{(x+2)^2(x-2)^2}$

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

$\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

$\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}$

$\implies \dfrac{7}{16}$

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0

3 years ago

What is the estimate of 1/2+1/6?

erastovalidia [21]

<span>2 — = 0.66667 3 hello

</span>

6 0

2 years ago

Read 2 more answers