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lapo4ka [179]
3 years ago
13

PLEASE ANSWER! ☹️ and round the answer to nearest whole number

Mathematics
1 answer:
Ksju [112]3 years ago
6 0
You can't round to the nearest whole number because you got to stay the same
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Find the product and sum of the roots of the equation:<br> 2x^2-9x-10=0
BabaBlast [244]

Answer:

<h3>           sum of the roots:    x_1+x_2=4\frac12</h3><h3>           product of the roots:     x_1x_2=-5             </h3><h3> Step-by-step explanation:</h3>

2x^2-9x-10=0\quad\implies\quad a=2\,,\ b=-9\,,\ c=-10

b^2-4ac=(-9)^2-4(2)(-10)=81+80=161>0

From Vieta's formulas applied to quadratic polynomial we have:

if    b^2-4ac\geqslant0   then

sum of roots:    x_1+x_2=-\dfrac ba=-\dfrac{-9}2=4\frac12

product of the roots:     x_1x_2=\dfrac ca=\dfrac{-10}2=-5

3 0
3 years ago
Which figure shows a reflection of ​pre-image QRS​ over the line t?
Igoryamba

Answer:

A, because its reflecting off of the y axis

Step-by-step explanation:

I have proof links, and I took the test

4 0
3 years ago
Find the sum of the first 6 terms of the sequence: –3, 1, 5, 9, . . .A) 45B) 42C) 48D) 25
Leni [432]

SOLUTION

From the sequence give

–3, 1, 5, 9, . . .

The first term, a = -3

The common difference, d = 4 (gotten by adding 4 to the next term).

The number of terms required n = 6.

Formula for sum of an arithmetic sequence is given by

S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack

Substituting these values into the equation above we have

\begin{gathered} S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack \\ S_6=\frac{6}{2}\lbrack2\times-3+(6-1)4\rbrack \\ S_6=3\lbrack-6+(5)4\rbrack \\ S_6=3\lbrack-6+20\rbrack \\ S_6=3\lbrack14\rbrack \\ S_6=42 \end{gathered}

Hence, the answer is 42, option B

3 0
1 year ago
How do you solve for the quotient of (x^-1) - 1 ÷ x - 1?
Ber [7]

\bf x^{-1}-1\div x-1\implies \implies \cfrac{1}{x}-1\div x-1\implies \cfrac{\frac{1}{x}-1}{~~x-1~~}\implies \cfrac{~~\frac{1-x}{x}~~}{\frac{x-1}{1}} \\\\\\ \cfrac{1-x}{x}\cdot \cfrac{1}{x-1}\implies \cfrac{-(\begin{matrix} x-1 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix})}{x}\cdot \cfrac{1}{\begin{matrix} x-1 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}\implies -\cfrac{1}{x}\implies -x^{-1}

3 0
4 years ago
Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?
likoan [24]

\mathrm{Use\:the\:rational\:root\:theorem}

a_0=15,\:\quad a_n=2

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5

=\left(x+3\right)\left(2x^2-11x+5\right)

Factor: 2x^2-11x+5

2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)

=x\left(2x-1\right)-5\left(2x-1\right)

2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)

\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:}

thus zeros of f(x) is

x=-3,\:x=5,\:x=\frac{1}{2}

5 0
3 years ago
Read 2 more answers
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