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3 years ago

What is 5m + 5n where m = 10 and n = -5?

Mathematics

1 answer:

patriot [66]3 years ago

8 0

Answer:

Step-by-step explanation:

5 times 10= 50

5 times -5= -25

50-25=25

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Three quarter of the students running a 100-yard race finished with an average time of 16 seconds. The remaining 25% of students

iren [92.7K]

Answer:

15 seconds

Step-by-step explanation:

Because the split id 25% and 75%, we could create another average pretending that there are four kids, one who ran in 12 seconds, and three who ran in 16.

Equation for averages: (a₁ + a₂ + a₃ + ... $a_{n}$ )/ n

Plug in:<em> (12 + 16 + 16 + 16)/4</em>

Add: 60/4

Divide: 15 seconds

5 0

3 years ago

Midpoint of (-2,-4) and (2,1)

Mice21 [21]

I got it =4/5 hope it’s right

7 0

3 years ago

A statement of an employee's biweekly earnings is given below. Earnings Deductions Week Ended Regular FED. SOC. MED STATE WITH.

DiKsa [7]

Answer: d. $852.96

Step-by-step explanation:

Given: The net pay = $667.17

Since net pay is the amount of money your employees take home after all deductions have been taken out.

We know that gross pay is the amount of money that employees receive before any taxes and deductions are taken out.

Thus to find the gross pay, we need to add all of the deductions to the net pay as:

Gross pay $=667.17+98+52.88+12.37+22.54 =\$852.96$

Hence, D is the right option.

8 0

3 years ago

Read 2 more answers

A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner

frozen [14]

Answer:

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Step-by-step explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner $I_{corner}$ is:

$I_{corner} = \int\limits \int\limits_R (x^2+y^2) \rho d A \\ \\ I_{corner} = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx$

where :

(a and b are the length and the breath of the rectangle respectively )

$I_{corner} = \rho \int\limits^a_0 {x^2y}+ \frac{y^3}{3} |^ {^ b}_{_0} \, dx$

$I_{corner} = \rho \int\limits^a_0 (bx^2 + \frac{b^3}{3})dx$

$I_{corner} = \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}$

$I_{corner} = \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]$

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Thus; the moment of inertia of the lamina about one corner is $I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

7 0

3 years ago

The height of a tree trunk is 20 meters and the base diameter is 0.5 meter.

Temka [501]

Question a:

Mass = Density × Volume
Density = Mass/Volume

Volume of the tree trunk (the shape of Cylinder) = Area of circular base × height
Volume = [πr²] × h
Volume = [π × 0.25²] × 20
Volume = 3.93 m³

Density = 380 kg/m³

Mass = Density × Volume
Mass = 380 × 3.93
Mass = 1493.4 kg

------------------------------------------------------------------------------------------------------------

Question b)

The growth ring = 4 millimeters = 4÷1000 = 0.004
New diameter = 0.5 + 0.004 = 0.5004
New height = 20 + 0.2 = 20.2

New volume = [πr²] × h
New volume = [π × 0.2502²] × 20.2
New volume = 3.97 m³

5 0

3 years ago