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Oxana [17]
3 years ago
7

What is the square root of 1,563?

Mathematics
1 answer:
murzikaleks [220]3 years ago
4 0
39.5 but if you round than it would be 40
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Plsss I need help!!!
mrs_skeptik [129]

Answer:

POSSIBLY  2

Step-by-step explanation: wasnt the best at math but i hope u do good

3 0
3 years ago
Rewrite the following equation in slope-intercept form. 6x - 3y = 19​
tensa zangetsu [6.8K]

Answer:

y= 2x - 19/3

Step-by-step explanation:

6x-3y=19

3y=6x-19

y=2x- 19/3

slope: 2

y-intercept: -19/3

7 0
3 years ago
Algebra help?<br><br> 343 = (x/6) ^2/3
Vladimir79 [104]

I can only assume that you meant, "Solve for x:"

Apply the exponent 3/2 to both sides of this equation.  The result will be

     3/2

343 = x/6.    

Multiplying both sides by 6 isolates x:

          3/2

6*343           = x                         Since 7^3 = 343,  the expression for x

                                                  can be rewritten as

           3/2

6*(7^3)          =   x                      which can be further simplified, as follows:


x =  6^(3/2)*7^(9/2), or:

x = 6^(3/2)*7^(8/2)*√7, or

x = 6^(3/2)*7^4*√7

3 0
3 years ago
Need the complete Answer of this problemssssss.
Mariana [72]
50 \times 0.90 = 45
Multiply 50 by 0.90 because you take a discount of 10% off. 1.00 = 100% 0.10 = 10% 0.01 = 1% if you multiply 50 by 0.10 instead you get the amount that is discounted instead of the new selling price.
5 0
3 years ago
An ice cube is melting, and the lengths of its sides are decreasing at a rate of 0.8 millimeters per minute At what rate is the
julia-pushkina [17]

Answer:

The rate of decrease is: 43.2mm^3/min

Step-by-step explanation:

Given

l = 18mm

\frac{dl}{dt} = -0.8mm/min ---- We used minus because the rate is decreasing

Required

Rate of decrease when: l = 18mm

The volume of the cube is:

V = l^3

Differentiate

\frac{dV}{dl} = 3l^2

Make dV the subject

dV = 3l^2 \cdot dl

Divide both sides by dt

\frac{dV}{dt} = 3l^2 \cdot \frac{dl}{dt}

Given that: l = 18mm and \frac{dl}{dt} = -0.8mm/min

\frac{dV}{dt} = 3 * (18mm)^2 * (-0.8mm/min)

\frac{dV}{dt} = 3 * 18 *-0.8mm^3/min

\frac{dV}{dt} = -43.2mm^3/min

<em>Hence, the rate of decrease is: 43.2mm^3/min</em>

8 0
3 years ago
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