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Paul [167]
3 years ago
11

What is 3/4 times 1/8=

Mathematics
2 answers:
vagabundo [1.1K]3 years ago
7 0

Answer: 3/32

Step-by-step explanation: When we multiply two fractions, such as 3/4 x 1/8, we simply multiply the numerators by each other and the denominators by each other.

lubasha [3.4K]3 years ago
6 0

Answer:

The answer should be 3/22.

Step-by-step explanation:

To multiply the fractions, multiply the numerators and denominators separately.

3 times 1 over 4 times 8,

Multiply straight across,

Answer should be 3/22

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Answer:

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Step-by-step explanation:

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Look at the image please help I’m struggling I have 5 more minutes to this test
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Wouldn’t it be $1.50 since half of 3 is 1.5
5 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}}
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How do you write 96% as a fraction in simplest form?
Troyanec [42]

Your answer would be 24/25

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How to find minimum and maximum of this equation.
Westkost [7]

Using it's vertex, the maximum value of the quadratic function is -3.19.

<h3>What is the vertex of a quadratic equation?</h3>

A quadratic equation is modeled by:

y = ax^2 + bx + c

The vertex is given by:

(x_v, y_v)

In which:

  • x_v = -\frac{b}{2a}
  • y_v = -\frac{b^2 - 4ac}{4a}

Considering the coefficient a, we have that:

  • If a < 0, the vertex is a maximum point.
  • If a > 0, the vertex is a minimum point.

In this problem, the equation is:

y + 4 = -x² + 1.8x

In standard format:

y = -x² + 1.8x - 4.

The coefficients are a = -1 < 0, b = 1.8, c = -4, hence the maximum value is:

y_v = -\frac{1.8^2 - 4(-1)(-4)}{4(-1)} = -3.19

More can be learned about the vertex of a quadratic function at brainly.com/question/24737967

#SPJ1

3 0
2 years ago
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