All categories

3 years ago

M(3-4m)=7+4(8-m2) m=?

Mathematics

1 answer:

mina [271]3 years ago

3 0

$m(3-4m)=7+4(8-m^2) \\
3m-4m^2=7+32-4m^2 \\
3m-4m^2+4m^2=7+32 \\
3m=39 \\
m=\frac{39}{3} \\
m=13$

You might be interested in

Your class wants to know to most typical age of student in your class. It collects the ages of all students and the teacher. Whi

dem82 [27]

Answer:

Median.

Step-by-step explanation:

Assuming the teacher is not of similar age to the students, their higher age would pull the mean higher than if the teacher were not included, where the median would not be pulled by this extreme value.

3 0

3 years ago

Estimate the difference of 422-284

Vlada [557]

The answer for this question is 138

8 0

3 years ago

Read 2 more answers

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Natasha2012 [34]

Answer:

$x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$ are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

$2x^2-10x-3$

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

$x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}$

General form of quadratic equation is $ax^2+bx+c$

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

$D=\sqrt{(-10)^2-4(2)(-3)}$

$\Rightarrow D=\sqrt{100+24}=\sqrt{124}$

Now substituting D in $x=\frac{b^2\pm\sqrt{D}}{2a}$ we get

$x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}$

$x=\frac{100\pm\sqrt{124}}{4}$

$x=\frac{100\pm2\sqrt{31}}{4}$

$x=\frac{50\pm\sqrt{31}}{2}$

Therefore, $x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$

5 0

3 years ago

Read 2 more answers

Answer this question please ASAP !!

harkovskaia [24]

Step-by-step explanation:

the median for the upper quartile is 21 or second quartile

8 0

3 years ago

Jackie is buying crackers and wants to know how many crackers are in the box. The box is 8 cm long, 2.4 cm wide, and 60 cm high.

Alja [10]

Answer:

Step-by-step explanation:

Number of crackers that can be fit into the box = volume of the box / volume of crackers

Volume = length x width x height

volume of the box = 8 cm x 2.4 cm x 60 cm = 1152 cm^3

Volume of the crackers = 3cm x 4 cm x 5cm = 60cm^3

1152/60 = 19.2

Rounding 19.2 to the nearest whole number gives 19

6 0

3 years ago