All categories

3 years ago

How do you find the sum of 74.365 and 9.82?

1 answer:

Citrus2011 [14]3 years ago

8 0

You have to add them, 74.365 + 9.82 = 84.185
You do it just like a normal addition problem but with decimals, the extra 5 at the end doesn’t matter just add another zero at the end of 9.820

You might be interested in

Could someone help me with these problems, please :)

Elanso [62]

Answer:

A for first one and C

Step-by-step explanation:

4 0

3 years ago

I need help. I’ll seriously make you brainly

ser-zykov [4K]

Answer:

a = 12

Step-by-step explanation:

add 2 to each side

5a/3 = a/4 + 17

distribute 12 to remove fractions

20a = 3a + 204

17a = 204

a = 12

7 0

3 years ago

If g(x) equals 4x + 5, which statement is true? Select the correct answer.

max2010maxim [7]

C is true as g intercept at 5 and f at 4.5

7 0

3 years ago

Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2

Finger [1]

Answer:

$\frac{y}{x^2}=\sin x+\pi$

Step-by-step explanation:

Consider linear differential equation $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$

It's solution is of form $y\,I.F=\int I.F\,q(x)\,dx$ where I.F is integrating factor given by $I.F=e^{\int p(x)\,dx}$ .

Given: $\frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x$

We can write this equation as $\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x$

On comparing this equation with $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$ , we get $p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x$

I.F = $e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}$ { formula used: $\ln a^b=b\ln a$ }

we get solution as follows:

$\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C$

{ formula used: $\int \cos x\,dx=\sin x$ }

Applying condition: $y(\pi)=\pi^2$

$\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C$

So, we get solution as :

$\frac{y}{x^2}=\sin x+\pi$

4 0

3 years ago

find continued product of (x+y) (x-y) (<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}"

leonid [27]

Answer:

$\large\boxed{(x+y)(x-y)(x^2+y^2)(x^4+y^4)=x^8-y^8}$

Step-by-step explanation:

$(x+y)(x-y)(x^2+y^2)(x^4+y^4)\\\\\text{use}\ (a-b)(a+b)=a^2-b^2\qquad(*)\\\\=\underbrace{(x+y)(x-y)}_{(*)}(x^2+y^2)(x^4+y^4)\\\\=\underbrace{(x^2-y^2)(x^2+y^2)}_{(*)}(x^4+y^4)\\\\=\left((x^2)^2-(y^2)^2\right)(x^4+y^4)\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\underbrace{(x^4-y^4)(x^4+y^4)}_{(*)}\\\\=\left(x^4\right)^2-\left(y^4\right)^2=x^8-y^8$

7 0

3 years ago