Plz this is important number 2 plz

Mathematics

1 answer:

ololo11 [35]3 years ago

7 0

Rounding it to the tens place so your answer for 12mm is going to be 10

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Guys I really need help with these too thanks u before!!!!

Natali5045456 [20]

Hey!!

here is your answer >>>

1) 3x - 10 > 11

let's take this as 3x - 10 = 11

3x = 21

x = 7
So, The numbers of X are always more than 7

2) 9x + 2 < 5

9x + 2 = 5

9x = 3

x = 1÷2

The numbers of X are always less than 1 ÷ 2

3) 2x + 8 < 6

2x + 8 = 6

2x = 2

x = 1

The numbers of X are less than 1

4) 4x - 3 < 12

4x - 3 = 12

4x = 15

x = 15 ÷ 4

The numbers of X are less than 15 ÷ 4

5) 2 ( 9 + x ) > 20

2 ( 9 + x ) = 20

9 + x = 10

x = 1

The numbers of X are more than 1

We can solve the rest using this

Hope my answer helps!

5 0

3 years ago

Translate the sentence into an equation.

gayaneshka [121]

4B+6=8

Step-by-step explanation:

4B+6=8

...

6 0

2 years ago

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

2 years ago

Math Help? 15 Points

frosja888 [35]

A last week Rosemary walked at a faster pace

7 0

3 years ago

Pls help???????? need to have an explanation also

Ratling [72]

Answer:

I would say the answer is 40 cookies

Step-by-step explanation:

The reason I say this is because if you write out the amount of cookies she can make with 6 cups of flour, you would get 30/6. 30 and 6 are both divisible by 6, so if you divided both by 6 you will get 5/1. 1 times 8 is 8, so if you multiply 5 by 8 you will get 40 cookies.

Hope this helped!

7 0

3 years ago