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4 years ago

What is the square root of sixty four

Mathematics

1 answer:

sweet [91]4 years ago

7 0

The square root of 64 is 8. 8 x 8 = 64 .

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3 1183 long division

Triss [41]

= 394 R 1

= 394 1/3

1183 divided by 3 equals
394 with a remainder of 1

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3 years ago

1. A triangle has two angles that measure 50° and 40°. The third angle is not labeled yet. Mark each of the

Shtirlitz [24]

Answer:

all sides have different lengths

and the third angle measures 90⁰

7 0

3 years ago

Help me please i really don’t get this

LekaFEV [45]

Answer:

The answer is D

Step-by-step explanation:

Coplanar means that they are on the same closed area(or plane)- P, M, C, N are all on the same Plane.

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3 years ago

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Given P space equals space P subscript 0 space plus space rho g h, your objective is to determine the uncertainty of P by measur

son4ous [18]

Answer:

the g's contributing term for the overall uncertainty of P is $dP_g = [\frac{dg}{g}]$

Step-by-step explanation:

From the question we are told that

The pressure is $P = P_o + \rho gh$

The first step in determining the uncertainty of P in by obtaining the terms in the equation contributing to it uncertainty and to do that we take the Ln of both sides of the equation

$ln P = lnP_o + ln(\rho gh )$

=> $ln P = lnP_o + ln \rho + ln g + ln h$

Then the next step is to differentiate both sides of the equation

$\frac{d(ln P)}{dP} = \frac{d(ln P_o)}{dP_o} + \frac{d(ln \rho)}{d\rho} +\frac{d(ln g)}{dg} + \frac{d(ln h)}{dh}$

=> $\frac{dP}{P} = \frac{dP_o}{P_o} + \frac{d \rho}{\rho} +\frac{d g}{g} + \frac{d h}{h}$

We asked to obtain the contribution of the term g to the uncertainty of P

This can deduced from the above equation as

$dP_g = [\frac{dg}{g}] P$

8 0

4 years ago

Please help will give brainliest

Ostrovityanka [42]

Answer:

1、 $\frac{-26}{1}$ =-26

2、-4

3、 $\frac{11}{2}$

4、 $-\frac{8}{11}$

5、At direct x equals 17, there's no slope

6、0

7、 $\frac{15}{14}$

8、 $-\frac{12}{30} =-\frac{2}{5}$

Step-by-step explanation:

In point-slope form, it takes about 3 seconds per problem

7 0

2 years ago

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