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3 years ago

Which is equivalent to (-m)4x n2 ?

Mathematics

2 answers:

Anna007 [38]3 years ago

6 0

Answer:

a.) m⁴n²

Step-by-step explanation:

yea

Tema [17]3 years ago

5 0

Answer:

a.) m⁴n²

Step-by-step explanation:

( -m)⁴ × n ²

A negative base raised to an even powers equals a positive.

m ⁴ × n²

multiply the terms

m⁴n²

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Five bags of chips cost $7.

Dvinal [7]

Answer:

1. $1.40 2. $8.40

Step-by-step explanation:

1. 7 divided by 5 is 1.4

2. 1.40 plus $7 equals $8.40

6 0

3 years ago

Read 2 more answers

4: A school has 314 boys & 310 girls. If they are to be grouped into equal classes of 26 each , how many classes will there

kherson [118]

4: A school has 314 boys & 310 girls. If they are to be grouped into equal classes of 26 each , how many classes will there be?

314 boys + 310 girls = 624

If they are grouped into classes of 26 students each, how many class will there be?

130 students = 5 classes

Therefore

260 students = 10 classes

260 x 2 = 520 = 20 classes

26 x 24 = 624

Number of classes = 24

Question 5).

Jim worked 48 weeks last year. Each week he worked 38 hours. If he worked an additional 240 hours of overtime, how many hours did he work in all?

Solution:

48 weeks

24hrs × 7 = 168hrs

168 x 48 = 8,064

Additional 240hrs

= 168hrs = 7 days (a week)

240hrs - 168 = 72 (3days)

168 + 72

= 240hrs + 8, 064

= 8, 304 hrs

6 0

3 years ago

A square + b square =

dmitriy555 [2]

c square

its making me write 20 more characters

ignore this

the answer is c squared

8 0

4 years ago

Read 2 more answers

Albert borrowed $12,000 at a

Aleks04 [339]

Answer:

the answer is A

Step-by-step explanation:

8 0

3 years ago

Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

Solution:

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

$x\left(x^{2}+7 x+12\right)=0$ ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

Therefore, the two roots are:

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0

3 years ago