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3 years ago

9

I am greater than 310. My number has 0 ones.The digits in my hundreds and tens places are the same

1 answer:

mixas84 [53]3 years ago

7 0

Step-by-step explanation:

Numbers: 330, 440, 550, 660 and so on all the way to 990

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What is the given point on the line of this equation y +2 = 5/3(x-3)

8_murik_8 [283]

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6

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3 years ago

Eight less than three times a number is -23

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3 years ago

What is the value of 1/2 + (-5/6) + 1/4?<br> 1. -7/12<br> 2. -1/12<br> 3. 1/12<br> 4. 1 7/12

algol [13]

Answer:

The answer is 4

Step-by-step explanation:

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2 years ago

(8n^2+4n^4)+(7n^3-4n^4)

Lina20 [59]

Answer:

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Step-by-step explanation:

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3 years ago

What is the other square root of 119 + 120i?

MakcuM [25]

Answer:

$\sqrt{119+120 i}=\pm (12.24+i 4.90)$

Step-by-step explanation:

Given

z = 119 + 120 i

Let $\sqrt{119+120 i}=p+iq$

Squaring both sides

$119+120 i=p^2-q^2+2ipq$

Comparing real and imaginary part

Re(LHS)=Re(RHS)

$119=p^2-q^2$ ...........................(1)

comparing Im(LHS)=Im(RHS)

120=2pq

$q=\frac{60}{p}$

Substitute q in 1

$119=p^2-(\frac{60}{p})^2$

$p^4-119p^2-(68)^2=0$

Let $x=p^2$

$x^2-119x-4624=0$

$x=\dfrac{119\pm \sqrt{119^2+4\times 4624}}{2}$

$x=\frac{119\pm 180.71}{2}$

we take only Positive value because $p^2=x$

x=149.85

$p^2=149.85$

thus $p=\pm 12.24$

$q=\pm 4.90$

thus,

$\sqrt{119+120 i}=\pm (12.24+i 4.90)$

8 0

3 years ago