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3 years ago

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Mathematics

2 answers:

8_murik_8 [283]3 years ago

8 0

Answer:

i think its D but im not exactly sure

LenaWriter [7]3 years ago

7 0

Answer:

Step-by-step explanation:

Subtract 60 from 80 to get 20, and divide 20 by 80 to get 1/4, which can be converted to 25%.

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Uca is designing a propeller system that consists of 3 blades, each of which is

zmey [24]

Answer:

The weight of the three blades together is $134\ g$

Step-by-step explanation:

step 1

Find the area of the circle

The area of the circle is equal to

$A=\pi r^{2}$

we have

$r=4\ in$

assume

$\pi=3.14$

substitute

$A=(3.14)(4)^{2}$

$A=50.24\ in^{2}$

step 2

Find the area of the 3 blades

we know that

The area of the 3 blades represent a central angle of 60 degrees, so

its a 1/6 of the area of complete circle

$A=50.24/6=8.37\ in^{2}$

step 3

Find the weight of the three blades together to the nearest gram

$(8.37\ in^{2})*(16\ g/in^{2})=134\ g$

4 0

3 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

3 years ago

What can 14/6 go into it’s simplest form? See

ladessa [460]

Answer:

7/3

Step-by-step explanation:

You divide 14 and 6 by 2 that is equal 7/3

6 0

3 years ago

Read 2 more answers

On Melissa's 6th birthday, she gets a $2000 CD that earns 7% interest compounded quarterly. If the CD matures on her 13th birthd