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zhuklara [117]
3 years ago
5

What the polynomial in standard form of (X-3)(X-4)=

Mathematics
2 answers:
IRINA_888 [86]3 years ago
6 0

///////////////

good luck

Anna007 [38]3 years ago
4 0

Answer:X^2-7x-12

Step-by-step explanation:

That’s how it is

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Can somebody help me pleaseeeee <br><br><br> Multiply -5/8•(-3)
Mars2501 [29]

Answer:3 [ 1 5 -5 6 0 0 ]

3 x 1 = 3

3 x 5 = 15

3 x -5 = -15

3 x 6 = 18

3 x 0 = 0

3 x 0 = 0

[ 3 15 -15 18 0 0 ]

[ 3 15 ]

[ -15 18 ]

[ 0 0]

The answer is B and I hope I explained this well for you

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
1/2x + 1/3y = 7
pashok25 [27]

let's multiply both sides in each equation by the LCD of all fractions in it, thus doing away with the denominator.

\begin{cases} \cfrac{1}{2}x+\cfrac{1}{3}y&=7\\\\ \cfrac{1}{4}x+\cfrac{2}{3}y&=6 \end{cases}\implies \begin{cases} \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{6}}{6\left( \cfrac{1}{2}x+\cfrac{1}{3}y \right)=6(7)}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( \cfrac{1}{4}x+\cfrac{2}{3}y\right)=12(6)} \end{cases}\implies \begin{cases} 3x+2y=42\\ 3x+8y=72 \end{cases} \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{using elimination}}{ \begin{array}{llll} 3x+2y=42&\times -1\implies &\begin{matrix} -3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~-2y=&-42\\ 3x+8y-72 &&~~\begin{matrix} 3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+8y=&72\\ \cline{3-4}\\ &&~\hfill 6y=&30 \end{array}} \\\\\\ y=\cfrac{30}{6}\implies \blacktriangleright y=5 \blacktriangleleft \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{substituting \underline{y} on the 1st equation}~\hfill }{3x+2(5)=42\implies 3x+10=42}\implies 3x=32 \\\\\\ x=\cfrac{32}{3}\implies \blacktriangleright x=10\frac{2}{3} \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left(10\frac{2}{3}~~,~~5 \right)~\hfill

7 0
3 years ago
In the paper drive, 4 boys brought old papers to school as follows: Fred, 30 pounds; Albert, 40 pounds; Henry, 10 pounds; and Pe
seraphim [82]
B: 30 pounds is the correct answer
5 0
3 years ago
Read 2 more answers
Which of the following are solutions to the equation below?
sladkih [1.3K]

Answer:

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

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

Step-by-step explanation:

considering the equation

2x^2\:-\:4x\:-\:3\:=\:x

solving

2x^2\:-\:4x\:-\:3\:=\:x

2x^2-4x-3-x=x-x

2x^2-5x-3=0

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

solving

x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

x=\frac{5+\sqrt{49}}{2\cdot \:2}

x=\frac{5+7}{4}

x=3

also solving

x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

x=\frac{5-\sqrt{49}}{4}

x=-\frac{2}{4}

x=-\frac{1}{2}

Therefore,

                 \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

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

7 0
3 years ago
Which fraction is not equivalent to 9/12<br> 15/20<br> 24/32<br> 6/8<br> 16/24
s2008m [1.1K]

Answer:

16/24

Step-by-step explanation:

9/12–> simplify to 3/4

3/4=15/20

24/32=3/4

6/8=3/4

16/24 is not equal

6 0
2 years ago
Read 2 more answers
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