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3 years ago

15

Find the value of given expression

qrt{8100} " alt=" \sqrt{8100} " align="absmiddle" class="latex-formula">

2 answers:

Andrew [12]3 years ago

8 0

90 is the answer for the question

Doss [256]3 years ago

6 0

Answer:

90 is the answers for the question

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How wide is a rectangle with a length of 38 cm and 8 dm perimeter?

Lesechka [4]

Perimeter is all four sides added together, therefore....

38+38= 76

8dm= 80cm
80cm*2=160 cm

76+160= 236<span />

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6<br> The quotient of 20 and 47.

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Answer:

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Step-by-step explanation:

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What is -2x - 10 = - 14x - 7?

jek_recluse [69]

x=4 because if you add 7 to 10 you get -2x-3=-14x then you add -2x to -14x and you get -3=-12 then you divide by -3 and -12 to get 4

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3 years ago

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Can you answer this question as soon as possible.

Whitepunk [10]

Answer:

The intensity of the first earthquake is 6.31 times the second one

Step-by-step explanation:

The given formulae is M=log( $\frac{I}{S}$ )

Here for both the earthquakes the value of S is same.

Let the Intensity of the first earthquake be I1 and the second be I2.

Given M1=7.5

M2=6.7

Now M1=log( $\frac{I1}{S}$ )------1

M2=log( $\frac{I2}{S}$ )-----2

Now subtract 1 and 2 we get

M1-M2=log( $\frac{I1}{I2}$ )

I1/I2= $10^{M1-M2}$

I1=I2( $10^{0.8}$

I1=I2*6.31

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3 years ago

Please help me I need help please

MakcuM [25]

Answer:

p = 4

Step-by-step explanation:

Given equation:

$x^2+(p-3)y^2-4x+6y-16=0$

<u>Standard equation of a circle:</u>

$(x-a)^2+(y-b)^2=r^2$

(where $(a,b)$ is the centre of the circle, and $r$ is the radius)

If you expand this equation, you will see that the coefficient of $y^2$ is always one.

Therefore, $p-3=1$

$\implies p=1+3=4$

<u>Additional information</u>

To rewrite the given equation in the standard form:

$\implies x^2+y^2-4x+6y-16=0$

$\implies x^2-4x+y^2+6y=16$

$\implies (x-2)^2-4+(y+3)^2-9=16$

$\implies (x-2)^2+(y+3)^2=16+4+9$

$\implies (x-2)^2+(y+3)^2=29$

So this is a circle with centre (2, -3) and radius √29

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2 years ago