Solve the system of linear equations by substitution

Consider the expression 40+24 Find the greatest common factor of the two numbers and rewrite the expression using the distributi

Natalka [10]

Answer:

Greatest Common Factor of 40+24 is 8

Rewriting the expression using the distributive property we get 8(5+3) =8(5)+8(3)

Step-by-step explanation:

We need to F=find the greatest common factor of the two numbers 40+24 and rewrite the expression using the distributive property

Greatest Common Factor:

The greatest number that is divisible by both numbers is known as greatest common factor.

So, Greatest Common Factor of 40+24 is 8

Now, Taking 8 common we get: 8(5+3)

Distributive property:

It is written as: a(b+c) = ab+ac

So, we have 8(5+3) =8(5)+8(3)

Rewriting the expression using the distributive property we get 8(5+3) =8(5)+8(3)

5 0

3 years ago

6 times the difference of a number and 3 is more than 24

Viefleur [7K]

Answer:

the number is 8

Step-by-step explanation:

8-3=5

6•5=30>24

expression: 6(x-3)>24

4 0

3 years ago

List possible rational zeroes of the function. 3x^5-8x^4+6x^2+7x-12.

spayn [35]

First list all the positive and negative factors of the constant term in the expression: ±(1,2,3,4,6,12) these will be the values for "p"

Second list all the positive and negative factors of the leading coefficient:
±(1,3) these will be the values for "q"

Now list all the possible values of $\frac{p}{q}$ these will be the possible rational zeros of the polynomial function:
±( $\frac{1}{1} , \frac{1}{3} , \frac{2}{1} , \frac{2}{3} , \frac{3}{1} , \frac{3}{3}, \frac{4}{1} , \frac{4}{3} , \frac{6}{1} , \frac{6}{3} , \frac{12}{1} , \frac{12}{3}$ )

these can be reduced to the following list:
±(1, $\frac{1}{3}$ , 2, $\frac{2}{3}$ , 3, 4, $\frac{4}{3}$ , 6, 12

This list represents the possible rational zeros of the function. You can then use synthetic division to narrow down the actual roots of the function.

8 0

3 years ago

What equals 10 in multiplicattion??

Yuliya22 [10]

5×2=10 1×10 those two factors equal 10. O cant think of anymore

4 0

3 years ago

Read 2 more answers

Solve the given initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1

Yuri [45]

Let's check if the ODE is exact. To do that, we want to show that if

$\underbrace{(x+y)^2}_M\,\mathrm dx+\underbrace{(2xy+x^2-2)}_N\,\mathrm dy=0$

then $M_y=N_x$ . We have

$M_y=2(x+y)$
$N_x=2y+2x=2(x+y)$

so the equation is indeed exact. We're looking for a solution of the form $\Psi(x,y)=C$ . Computing the total differential yields the original ODE,

$\mathrm d\Psi=\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0$
$\implies\begin{cases}\Psi_x=(x+y)^2\\\Psi_y=2xy+x^2-2\end{cases}$

Integrate both sides of the first PDE with respect to $x$ ; then

$\displaystyle\int\Psi_x\,\mathrm dx=\int(x+y)^2\,\mathrm dx\implies\Psi(x,y)=\dfrac{(x+y)^3}3+f(y)$

where $f(y)$ is a function of $y$ alone. Differentiate this with respect to $y$ so that

$\Psi_y=2xy+x^2-2=(x+y)^2+f'(y)$
$\implies2xy+x^2-2=x^2+2xy+y^2+f'(y)$
$f'(y)=-2-y^2\implies f(y)=-2y-\dfrac{y^3}3+C$

So the solution to this ODE is

$\Psi(x,y)=\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3+C=C$

i.e.

$\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3=C$

6 0

3 years ago