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3 years ago

PLEASE HELP !! ILL GIVE 40 POINTS !!! PLUS BRAINLIEST !!

Mathematics

1 answer:

OLEGan [10]3 years ago

6 0

Answer:

QPN and RSU

Step-by-step explanation:

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PLZZZ NEED HELP!!!!!!

shtirl [24]

B is the correct answer

5 0

3 years ago

Read 2 more answers

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

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

Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

4 0

3 years ago

The area of a square can be found using the equation A = s², where A is the area and s is the measure of one side of the square.

Galina-37 [17]

Hi there!

Let's use the given formula of a square (A = s^2), to figure out what the equation would be. We are looking for one side, or s here.

WORK:
81 = s^2
squareroot(81) = s

ANSWER:
The second option, root 81.

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

3 0

4 years ago

Write the equation of a line in point-slope form that has a slope of -1 and passes through the point (-2, 5).

Angelina_Jolie [31]

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ -2}}\quad ,&{{ 5}})\quad 
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies -1
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad 
\begin{array}{llll}
\textit{plug in the values for }
\begin{cases}
y_1=5\\
x_1=-2\\
m=-1
\end{cases}\\
\textit{and solve for "y"}
\end{array}\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}$

8 0

3 years ago

Read 2 more answers

Can someone please help me

lesya [120]

The first one is no solution, the second one is t+x=y

5 0

3 years ago