All categories

2 years ago

25 - X= pls anwer this

Mathematics

2 answers:

OlgaM077 [116]2 years ago

5 0

Answer:

x = 25

Step-by-step explanation:

hope this helps

Alik [6]2 years ago

4 0

-x = -25
x=25
u take the 25 to the other side then divide both of them be -

You might be interested in

Find the global maximum and global minimum values (if they exist) of x 2 + y 2 in the region x + y = 1. If there is no global ma

Grace [21]

Given that $x+y=1$ , we have $y=1-x$ , so that

$f(x,y)=x^2+y^2\iff g(x)=x^2+(1-x)^2$

Take the derivative and find the critical points of $g$ :

$g'(x)=2x-2(1-x)=4x-2=0\implies x=\dfrac12$

Take the second derivative and evaluate it at the critical point:

$g''(x)=4>0$

Since $g''$ is positive for all $x$ , the critical point is a minimum.

At the critical point, we get the minimum value $g\left(\frac12\right)=f\left(\frac12,\frac12\right)=\frac12$ .

5 0

3 years ago

Is the histogram uniform, symmetric, or skewed?

hodyreva [135]

A histogram is a graph used to present ranges of data. In contrast to bar graphs or vertical graphs, these are more used in displaying complex and higher statistical data ranges. Moreover, the illustration is Uniform. Uniform in that it portrays that the data is in a rectangular shaped form.

8 0

4 years ago

Help with q25 please. Thanks.

Westkost [7]

First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.

Let's apply the first derivative of this f(x) function.

$f(x) = \sin(px)+\cos(px)\\\\f'(x) = \frac{d}{dx}[f(x)]\\\\f'(x) = \frac{d}{dx}[\sin(px)+\cos(px)]\\\\f'(x) = \frac{d}{dx}[\sin(px)]+\frac{d}{dx}[\cos(px)]\\\\f'(x) = p\cos(px)-p\sin(px)\\\\ f'(x) = p(\cos(px)-\sin(px))\\\\$

Now apply the derivative to that to get the second derivative

$f''(x) = \frac{d}{dx}[f'(x)]\\\\f''(x) = \frac{d}{dx}[p(\cos(px)-\sin(px))]\\\\ f''(x) = p*\left(\frac{d}{dx}[\cos(px)]-\frac{d}{dx}[\sin(px)]\right)\\\\ f''(x) = p*\left(-p\sin(px)-p\cos(px)\right)\\\\ f''(x) = -p^2*\left(\sin(px)+\cos(px)\right)\\\\ f''(x) = -p^2*f(x)\\\\$

We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.

Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.

-----------------------------------

Let's compute dy/dx. We'll use f(x) as defined earlier.

$y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\$

Use the chain rule here.

There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.

Now use the quotient rule to find the second derivative of y

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\$

If you need a refresher on the quotient rule, then

$\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\$

where P and Q are functions of x.

-----------------------------------

This then means

$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\$

Note the cancellation of -(f ' (x))^2 with (f ' (x))^2

------------------------------------

Let's then replace f '' (x) with -p^2*f(x)

This allows us to form ( f(x) )^2 in the numerator to cancel out with the denominator.

$\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*f(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*(f(x))^2}{(f(x))^2} + p^2\\\\-p^2 + p^2\\\\0\\\\$

So this concludes the proof that $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2 = 0\\\\$ when $y = \ln\left(\sin(px)+\cos(px)\right)\\\\$

Side note: This is an example of showing that the given y function is a solution to the given second order linear differential equation.

7 0

3 years ago

Write an equation of the circle given its center and radius

lapo4ka [179]

Answer:

x²+y²-4x-6y+12.75=0

Step-by-step explanation:

The equation is given by (x-h)²+(y-k)²=r²

center is (2,3), so h=2, k=3,r=0.5

equation is given by;

(x-2)²+(y-3)²=0.5²

Expand

x(x-2)-2(x-2)+y(y-3)-3(y-3)=0.5²

4-2x-2x+x²+9-3y-3y+y²=0.25

4-4x+x²+9-6y+y²=0.25

x²-4x+13+y²-6y=0.25

x²-4x+y²-6y+13=0.25

x²-4x+y²-6y+12.75=0

x²+y²-4x-6y+12.75=0

7 0

3 years ago

Read 2 more answers

Zher and Gabbie are making cookies for the Hospitable Pantry. They only have up to 6 hours to

GrogVix [38]

Answer:

32x3=96. 20x3= 60. 60 + 96=156 if you make 3 hours of chocolate chip and 3 hours snickerdoodles you will have 156 cookies

4 0

3 years ago