$\hspace{34em}|\\ \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{20\% of 56.43}}{\left( \cfrac{20}{100} \right)56.43}\implies 11.286~\hfill \stackrel{about}{\approx 11.29}$

6 0

3 years ago

Use synthetic division to find P (-10) for P(x)=2x^3+14x^2-58x

Savatey [412]

The polynomial remainder theorem states that the remainder upon dividing a polynomial $p(x)$ by $x-c$ is the same as the value of $p(c)$ , so to find $p(-10)$ you need to find the remainder upon dividing

$\dfrac{2x^3+14x^2-58x}{x+10}$

You have

..... | 2 ... 14 ... -58
-10 | ... -20 ... 60
--------------------------
..... | 2 ... -6 .... 2

So the quotient and remainder upon dividing is

$\dfrac{2x^3+14x^2-58x}{x+10}=2x-6+\dfrac2{x+10}$

with a remainder of 2, which means $p(-10)=2$ .

5 0

4 years ago

can someone help me!! Use algebraic rules of equations to predict the solution type to the system of equations. Include all of y

xz_007 [3.2K]

Answer:

x + y = 4 Equation 1

y = 2x -1 Equation 2

-2x + y = -1

x + y = 4

Subtract

You will get

- 3x = -5

Divide by -3

-3x / -3 = - 5 /-3

x = 5/3

Substitute x into equation 1

x + y = 4

x = 5/3

5/3 + y = 4

y = 4 - 5/3

y = 7/3

x = 5/3 y = 7/3

Hope this helps.

4 0

3 years ago

Find an equation for the plane through the points p=(0,0,0), q=(0,1,0), and r=(1,2,3)

jek_recluse [69]

Hello,

Assume Ax+By+Cz+D=0 the plane 's equation.

p=(0,0,0)==>A*0+B*0+C*0+D=0==>D=0

q=(0,1,0)==>A*0+B*1+C*0+0=0==>B=0

r=(1,2,3)==>A*1+0*2+C*3+0=0==>A=-3C
Let C=1==>A=-3

An equation of the plane is -3x+z=0

6 0

4 years ago