All categories

3 years ago

EMERGENCY!!!! I have fallen and can't get up.

Mathematics

2 answers:

Novay_Z [31]3 years ago

8 0

Answer:

do you need a bandaid?

Step-by-step explanation:

GaryK [48]3 years ago

4 0

Step-by-step explanation:

Wth?

Are u okay??

call 911

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Devin was helping his grandfather gather corn in the

SVETLANKA909090 [29]

Answer:

$4/hour

Step-by-step explanation:

From 7:30 am to 10:00 am it's 2.5 hours.

10 * 2.5 hours = 25 hours

He worked a total of 25 hours.

$100/(25 hours) = $4/hour

7 0

4 years ago

Can someone please please help me out I don’t understand this

MatroZZZ [7]

Answer:

Step-by-step explanation:

5 0

3 years ago

Solve using a formula. Please don't guess, I would like professionals to take a look

Marianna [84]

Let's take this problem step by step:

What we know:

$x+y=4\\xy=-2$

Before we solve, let's do one thing that will help us out greatly later down the road:

$x+y=4\\(x+y)^2=4^2\\x^2+2xy+y^2=16\\x^2+2(xy)+y^2=16\\x^2+2(-2)+y^2=16\\x^2+4+y^2=16\\x^2+y^2=20$ <--- useful equation

Let's rearrange the problem a little bit:

$x+\frac{x^3}{y^2}+\frac{y^3}{x^2} +y=\frac{x^3}{x^2} +\frac{x^3}{y^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}$

Combine fractions of common denominators:

$\frac{x^3+y^3}{x^2} +\frac{x^3+y^3}{y^2} =(x^3+y^3)*(\frac{1}{x^2}+\frac{1}{y^2} )$

Now's let factor everything apart:

$(x^3+y^3)=(x+y)(x^2-xy+y^2)\\\\\frac{1}{x^2}+\frac{1}{y^2} =\frac{x^2+y^2}{x^2y^2}$

Let's use what we know and our useful equation:

$(x+y)*(x^2-xy+y^2)*(\frac{x^2+y^2}{x^2y^2} )\\=4*(x^2+y^2-xy)*(\frac{20}{(xy)^2} )\\=4*(20-(-2))*\frac{20}{(-2)^2} \\=4*22*5\\=440$

The value is 440

Answer: 440

Hope that helps!

#LearnwithBrainly

6 0

2 years ago

Wxyz is a dilation image of wxyz. Wich is the correct description of the dilation?

garri49 [273]

It’s either bigger or smaller

3 0

2 years ago

Which point is the solution to the following system of equations?

Naya [18.7K]

The point (3, 2) is the solution to given system of equations

Solution:

Given that system of equations are:

$x^2 + y^2 = 13$ ------ eqn 1

$2x - y = 4$ ------- eqn 2

From eqn 2,

y = 2x - 4

Substitute y = 2x - 4 in eqn 1

$x^2 + (2x - 4)^2 = 13\\\\x^2 + 4x^2 + 16 - 16x = 13\\\\5x^2 -16x + 3 = 0$

Let us solve the above equation by quadratic formula,

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the Quadratic Formula for $5x^2 -16x + 3 = 0$ where a = 5, b = -16, and c = 3

$\begin{aligned}&x=\frac{-(-16) \pm \sqrt{(-16)^{2}-4(5)(3)}}{2 \times 5}\\\\&x=\frac{16 \pm \sqrt{256-60}}{10}\\\\&x=\frac{16 \pm \sqrt{196}}{10}\end{aligned}$

The discriminant $b^2 - 4ac>0$ so, there are two real roots.

$\begin{aligned}&x=\frac{16 \pm \sqrt{196}}{10}=\frac{16 \pm 14}{10}\\\\&x=\frac{16+14}{10} \text { or } \frac{16-14}{10}\\\\&x=\frac{30}{10} \text { or } x=\frac{2}{10}\\\\&x=3 \text { or } x=0.2\end{aligned}$

Substitute for x = 0.2 and x = 3 in 2x - y = 4

when x = 3

2(3) - y = 4

6 - y = 4

y = 2

when x = 0.2

2(0.2) - y = 4

0.4 - y = 4

y = 0.4 - 4

y = -3.6

Thus Option D is correct The point is (3, 2)

7 0

3 years ago