Help me with this question please

How do I solve this question?

Sati [7]

Answer:

m=2 and n=3

Step-by-step explanation:

Step :-

Given $[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4$

using algebraic formula $(a^m)^n= a^{mn}$

now

$2^{m} x^{mn} y^{2m} = 4x^{6} y^{4}$

now equating 'x' powers, we get

$x^{mn} = x^{6}$

$mn=6$ ....(1)

now

$y^{2m} = y^{4}$

Equating 'y' powers ,we get

2 m=4

m=2

substitute m= 2 in equation (1)

we get

2 n=6

n=3

verification:-

substitute m=2 and n=3 , we get

$[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4$

$(2 x^{3} y^2)^2 = 4 x^6 y ^4\\$

$4 x^6 y^4 = 4 x^6 y^4$

both are equating so m= 2 and n=3

4 0

2 years ago

Plot point C(2,7). If M is the midpoint of CD , what are the coordinates of D?

Y_Kistochka [10]

Answer:

The coordinates of D is (-4,-9)

Step-by-step explanation:

Given

C = (2,7)

M=(-1,-1) ----- Missing part of question

Required

Determine the coordinates of D

Let the coordinates of D be (x₂,y₂)

We'll solve this question using mid point formula;

$M(x,y) = (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Where

$C(x_1,y_1) = (2,7)$ and $M(x,y) = (-1,-1)$

Put these values in the formula

$(-1,-1) = (\frac{2+x_2}{2},\frac{7+y_2}{2})$

By comparison;

$-1 = \frac{2+x_2}{2}$ and $-1 = \frac{7+y_2}{2}$

$-1 = \frac{2+x_2}{2}$

Multiply both sides by 2

$-2 = 2 + x_2$

Subtract 2 from both sides

$-2 -2 = x_2$

$x_2 = -4$

$-1 = \frac{7+y_2}{2}$

Multiply both sides by 2

$-2 = 7 + y_2$

Subtract 7 from both sides

$-2 - 7 = y_2$

$y_2 = -9$

Hence, the coordinates of D is (-4,-9)

4 0

3 years ago

What is the area of this triangle in the coordinate plane? 5 units² 6 units² 7 units² 12 units² A graph with a triangle drawn be

____ [38]

Answer:

6 units²

Step-by-step explanation:

The coordinates (1,1) (1,4) (5,1) form a right triangle as shown in the graph below

The base is is (1,1) and (5,1) so the length is 5-1 or 4 units

The height is (1,1) and (1,4) so the height is 4-1 or 3 units

Area of a triangle is given by

A = 1/2 bh

= 1/2 (4*3)

= 1/2 *12

= 6 units^2

8 0

3 years ago

If the term of arthematic progression decrease in magnitude then common difference must be

Tpy6a [65]

We can say that if the term of arithmetic progression decrease in magnitude then the common difference must be negative.

An arithmetic progression is a sequence where every term is a sum of the previous term and a common difference.

If we suppose the first term of an arithmetic progression to be a, and the common difference to be d, then the arithmetic progression goes like this:

The first term, a₁ = a,

Second term, a₂ = a₁ + d = a + d,

Third term, a₃ = a₂ + d = a + 2d,

Fourth term, a₄ = a₃ + d = a + 3d, and it goes so on up to,

The n-th term, aₙ = aₙ₋₁ + d = a + (n - 1)d.

In the question, we are asked if the term of arithmetic progression decrease in magnitude then what can we say about the common difference.

As the magnitude of the term decreases, we can show that as:

The n-th term < The first term,

or, aₙ < a₁,

or, a + (n - 1)d < a,

or, a + (n - 1)d - a < a - a {Subtracting a from both sides},

or, (n - 1)d < 0.

Since, n > 1, (n - 1) > 0, thus we get the common difference, d < 0, that is, the common difference is negative.

Thus, we can say that if the term of arithmetic progression decrease in magnitude then the common difference must be negative.

Learn more about arithmetic progressions at

brainly.com/question/6561461

#SPJ4

8 0

2 years ago

Will mark brainliest!

AfilCa [17]

Answer:

Area under the graph = 6 sq.units

I've shown the complete calculation over the attached page

7 0

1 year ago