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3 years ago

15-12 + 4(3) = 15-12-12 15-1 14

Mathematics

2 answers:

MissTica3 years ago

7 0

Answer:

Step-by-step explanation:

irina [24]3 years ago

7 0

Answer:

Step-by-step explanation:

15-12+4(3)

15-12+12

3+12

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Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

3 years ago

Three times a number, minus 2, is equal to two times the number, plus 7.

WINSTONCH [101]

Answer:

3x - 2 = 2x + 7

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

SV is a midsegment of △RTU.<br> If TU=y+38 and SV=y–9, what is the value of y?

pav-90 [236]

Answer:

y = 56

Step-by-step explanation:

the midsegment SV is half the length of the side TU , that is

y - 9 = $\frac{1}{2}$ (y + 38) ← multiply both sides by 2 to clear the fraction

2y - 18 = y + 38 ( subtract y from both sides )

y - 18 = 38 ( add 18 to both sides )

y = 56

8 0

2 years ago

Is (x-1) a factor of 2x3 - 15x2 + 22x + 15

Advocard [28]

Answer:

Step-by-step explanation:

If x - 1 is a factor of the cubic, then 1 should result in 0 when you put it in the cubic for x. Does it?

f(1) = 2(1)^3 - 15(1)^2 + 22(1) + 15

f(1) = 2 - 15 + 22 + 15

f(1)= 24.

No the result is not zero. So x - 1 is not a factor.

However there must be something that makes this cubic go to zero. The easiest way to find it is to graph it.

The numbers that do make this zero are -0.5, 3, 5

Which mean that the factors are (2x+1)(x-3)(x - 5)

6 0

3 years ago

Need Help ASAP WILL MARK YOU BRAINLYEST

Dima020 [189]

he divided by 2 instead of taking the square root so the actual answer should be 26 and for the second the 20 is across from the 90 so its suppose to be c

the answer should be 16

5 0

3 years ago