Hurry Quick, help me do this problem. I don't understand how to do this so please help! (Problem in photo)

Mathematics

1 answer:

matrenka [14]3 years ago

7 0

Answer:

1) The system of equations is $5x+4y=75$ and $x+y=18$

2) The first number is $3$ and the second number is $15$

Step-by-step explanation:

1) Let be "x" the first number and "y" the second number.

Remember that:

a- The word "times" indicates multiplication.

b- A sum is the result of an addition.

c- "Is" indicates this sign: $=$

Then, the sum of 5 times "x" and 4 times "y" is 75, can written as:

$5x+4y=75$

And "The sum of the two numbers is 18" can written as:

$x+y=18$

Therefore, the System of equations is:

$\left \{ {{5x+4y=75} \atop {x+y=18}} \right.$

2) You can use the Elimination Method to solve it:

- Multiply the second equation by -5, add the equations and then solve for "y":

$\left \{ {{5x+4y=75} \atop {-5x-5y=-90}} \right.\\......................\\-y=-15\\\\y=15$

- Substitute the value of "y" into any original equation and solve for "x":

$x+15=18\\\\x=18-15\\\\x=3$

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Find the zero of each function and state the multiplicity of each zero. Please show all steps.

vodka [1.7K]

Answer:

1. y=(x+3)^3. Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1). Zeros: x=2 multiplicity 2; x=1 multiplicity 1.

3. y=(2x+3)(x-1)^2. Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

Step-by-step explanation:

1. y=(x+3)^3

$y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3$

Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1)

$y=0\\ (x-2)^2(x-1)=0\\ \left \{ {{(x-2)^2=0} \atop {x-1=0}} \right\\ \left \{ {{\sqrt{(x-2)^2} =\sqrt{0} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x-2=0} \atop {x=1}} \right\\ \left \{ {{x-2+2=0+2} \atop {x=1}} \right\\ \left \{ {{x=2} \atop {x=1}} \right.$

Zeros: x=2 multiplicity 2; x=1 multiplicity 1

3. y=(2x+3)(x-1)^2

$y=0\\ (2x+3)(x-1)^2=0\\ \left \{ {{2x+3=0} \atop {(x-1)^2=0}} \right\\ \left \{ {{2x+3-3=0-3} \atop {\sqrt{(x-1)^2} =\sqrt{0} }} \right\\ \left \{ {{2x=-3} \atop {x-1=0}} \right\\ \left \{ {{\frac{2x}{2} =\frac{-3}{2} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x=-\frac{3}{2} } \atop {x=1}} \right.$