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4 years ago

What is 1.58 in standard form?

Mathematics

1 answer:

ella [17]4 years ago

8 0

1 one and 58 hundredths.

I hope you like this answer, and have a good night and life! :D

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9p+3=-p+7+2p+3p what is the answers to this equation

Nadusha1986 [10]

Answer:

p= 4/5

Step-by-step explanation:

9p-3p-2p+p=7-3

5p= 4

p= 4/5

4 0

3 years ago

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When would u need an interpreter? What would that person do?

seropon [69]

If you speak a different language this person can interpret your speaking into the language understandable by the other party

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4 years ago

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Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

3 0

3 years ago

I need help with this one

Reika [66]

Answer:

look it up

Step-by-step explanation:

8 0

3 years ago

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F(x) = 5x^3+ 2x^2 - 90x - 36. <br><br> finding all zeros by factoring and explaining the steps

ella [17]

Answer:

f(x)=5x^3-2x^2-90x-36=0

=x^2(5x-2)-18(5x-2)=(x^2-18)(5x-2)=0

x^2-18=0/5x-2=0

x^2=18=x=9√2

5x-2=0

x=2/5

zeros are 9√2,2/5

8 0

3 years ago