Help me please please

Find the distance between the two given points. (-6, -5) and (2, 0)

Ipatiy [6.2K]

Answer:

The answer is

<h2> $\sqrt{41} \: \: or \: \: \: 6.403 \: \: \: units$ </h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{( {x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-6, -5) and (2, 0)

The distance is

$d = \sqrt{ ({ - 6 - 2})^{2} + ({ - 5 - 0})^{2} } \\ = \sqrt{ ({ - 4})^{2} + ({ - 5})^{2} } \\ = \sqrt{16 + 25} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

We have the final answer as

$\sqrt{41} \: \: or \: \: \: 6.403 \: \: \: units$

Hope this helps you

4 0

3 years ago

Joe earns $13.40 an hour and it's paid every two weeks. Last week he worked seven hours longer than the week before. His pay for

andre [41]

Answer: 51

Step-by-step explanation:

You divide both numbers

683.40 / 13.40

5 0

3 years ago

Tom asked his Granny how old she was. Rather than giving him a straight answer, she replied: "I have 6 children, and there are 4

Ede4ka [16]

Answer: 58 years old

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Please help me solve this problem ASAP

DiKsa [7]

$\bold{\huge{\blue{\underline{ Solution }}}}$

<h3>Given :-</h3>

The right angled below is formed by 3 squares A, B and C
The area of square B has an area of 144 inches ²
The area of square C has an of 169 inches ²

We have to find the area of square A?

<h3>Let's Begin :- </h3>

The right angled triangle is formed by 3 squares

We have, 

Area of square B is 144 inches²
Area of square C is 169 inches²

We know that,

$\bold{ Area \: of \: square = Side × Side }$

Let the side of square B be x

Subsitute the required values,

$\sf{ 144 = x × x }$

$\sf{ 144 = x² }$

$\sf{ x = √144}$

$\bold{\red{ x = 12\: inches }}$

Thus, The dimension of square B is 12 inches

Area of square C = 169 inches

Let the side of square C be y

Subsitute the required values,

$\sf{ 169 = y × y }$

$\sf{ 169 = y² }$

$\sf{ y = √169}$

$\bold{\green{ y = 13\: inches }}$

Thus, The dimension of square C is 13 inches.

It is mentioned in the question that, the right angled triangle is formed by 3 squares

The dimensions of square be is x and y

Let the dimensions of square A be z

<h3>Therefore, By using Pythagoras theorem, </h3>

The sum of squares of base and perpendicular height equal to the square of hypotenuse

That is,

$\bold{\pink{ (Perpendicular)² + (Base)² = (Hypotenuse)² }}$

Here, 

Base = x = 12 inches
Perpendicular = z
Hypotenuse = y = 13 inches

Subsitute the required values,

$\sf{ (z)² + (x)² = (y)² }$

$\sf{ (z)² + (12)² = (169)² }$

$\sf{ (z)² + 144 = 169}$

$\sf{ (z)² = 169 - 144 }$

$\sf{ (z)² = 25}$

$\bold{\blue{ z = 5 }}$

Thus, The dimensions of square A is 5 inches

<h3>Therefore,</h3>

Area of square

$\sf{ = Side × Side }$

$\sf{ = 5 × 5 }$

$\bold{\orange{ = 25\: inches }}$

Hence, The area of square A is 25 inches.

6 0

2 years ago

SHORT ANSWER

Zigmanuir [339]

Gradient of the line formed;
m= $\frac{change in y}{change in x}$
m= $\frac{-2-1}{6--5}$
m= $\frac{-3}{11}$
y=mx+c
y= $\frac{-3}{11} x+c$
Replacing for x and y;
1=-3/11*(-5)+c
1=15/11+c
c=1-15/11
c=-4/11
y= $\frac{-3}{11} x- \frac{4}{11}$

4 0

3 years ago