All categories

3 years ago

Please help, refer to the picture for information.

Mathematics

1 answer:

aleksklad [387]3 years ago

5 0

Answer:

Step-by-step explanation:

Using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 1, 4) and (x₂, y₂ ) = (4, 1)

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

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

= $\sqrt{25+9}$ = $\sqrt{34}$ → b

You might be interested in

a phone company offers two monthly charge, in Plan a, the customer pays a monthly fee of $40.10 and then an additional 4 cents p

Charra [1.4K]

Answer:

x > 1.7 minutes

The monthly telephone usage amounts so that plan A is not greater than plan B are all greater than 1.7 minutes.

Step-by-step explanation:

The two plans must be defined by the equation of the line y = mx + b, where

y = plan

m = slope or payment of additional cents per minute

x = time expressed in minutes

For Plan A, we have

y = 4x + 40.10 (Equation A)

While plan B is defined as

y = 7x + 35 (Equation B)

Plan A must be less than Plan B,

4x + 40.10 < 7x + 35

We put the “x” on the left side and the independent terms on the right side,

4x - 7x < 35 - 40.10

We add algebraically,

-3x < -5.10

We multiply the equation by -1 to eliminate the two “minus” signs, changing the inequality sign,

3x > 5.10

We isolate x,

x > 5.10 / 3

We solve, calculating the value of x,

x > 1.7 minutes

The monthly telephone usage amounts so that plan A is not greater than plan B are all greater than 1.7 minutes.

7 0

3 years ago

When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F

aleksandr82 [10.1K]

Let $x$ be the unknown number. So, three times that number means $3x$ , and the square of the number is $x^2$

We have to sum 528 and three times the number, so we have $528+3x$

Then, we have to subtract this number from $x^2$ , so we have

$x^2-(3x+528)$

The result is 120, so the equation is

$x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0$

This is a quadratic equation, i.e. an equation like $ax^2+bx+c=0$ . These equation can be solved - assuming they have a solution - with the following formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

If you plug the values from your equation, you have

$x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}$

So, the two solutions would be

$x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27$

$x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24$

But we know that x is positive, so we only accept the solution $x = 27$

6 0

3 years ago

Algebra Functions and Data Analysis

-Dominant- [34]

Hello there!

The correct answer is option C (-∞,10)

Best of luck with your study!

8 0

3 years ago

Whats 2099 dived by 800

sergejj [24]

Answer:

2099/800= 2.62375

Step-by-step explanation:

5 0

3 years ago

Can one of you guys help me please :( thank you

timurjin [86]

Answer:

42???

Step-by-step explanation:

sorry if wrong

hope this helps

5 0

3 years ago

Read 2 more answers