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attashe74 [19]
3 years ago
15

How do solve this?? please

Mathematics
1 answer:
aksik [14]3 years ago
6 0

Answer:

m = - 3

n = 2

Step-by-step explanation:

It can be solved using laws of exponents as given below:

5 {x}^{n} y(2 {x}^{4}  {y}^{m} ) =  \frac{10 {x}^{6} }{ {y}^{2} }  \\  \\ (5  {x}^{n} .2 {x}^{4} )(y. {y}^{m} ) = 10 {x}^{6}  {y}^{ - 2}  \\  \\\cancel {10}\:\: {x}^{n + 4} . {y}^{m + 1}  = \cancel {10}\:\: {x}^{6}  {y}^{ - 2}  \\  \\ {x}^{n + 4} . {y}^{m + 1}  =  {x}^{6}  {y}^{ - 2}  \\equating \: like \: terms \: on \: both \: sides  \\  \\  {x}^{n + 4}  =  {x}^{6}  \\  \\  \implies \: n + 4 = 6 \\  \\ n = 6 - 4 \\  \\  \huge \red{n = 2} \\  \\  {y}^{m + 1}  =  {y}^{ - 2}  \\  \\  \implies \: m + 1 =  - 2 \\  \\ m =  - 2 - 1 \\  \\  \huge \purple{m =  - 3}

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Hi Can someone help this is hard?
iren [92.7K]

The domain is the set of all real numbers and the range is the set of all real numbers larger than -4, so the correct option is the third one.

<h3 /><h3>How to get the domain and the range?</h3>

For a function y = f(x) we define the domain as the set of the x-values (horizontal axis) and the range as the set of the y-values (vertical axis).

Here we have a parabola, in this case the domain is always the set of all real numbers, and the range is all the values above or equal to the vertex (in cases like this, where the parabola opens up).

We can see that the vertex is at y = -4

So the range is.

[-4, ∞]

So the correct option is 3.

Learn more about domain and range:

brainly.com/question/10197594

#SPJ1

6 0
1 year ago
What is the square root of 25p^3 expressed in simplified form?
notka56 [123]
The first step to solving this is to factor out the perfect square of 5²
\sqrt{ 5^{2}  p^{3} }
next,, factor out the perfect square of p³
\sqrt{ 5^{2}  p^{2} x p}
the root of a product is equal to the product of the roots of each factor 
√5² √p² √p
now reduce the first index of the radical and the exponent with 2
5√p² √p
then reduce the second index of the radical and exponent with
5p√p
this means that the correct answer to your question is 5p√p
let me know if you have any further questions
:)
6 0
4 years ago
Find the value of c that makes x^2+26x+c a perfect-square trinomial.
MakcuM [25]
C = 169

square is (x + 13)^2

Explanation:

In the form:
ax^2+ bx +c
to complete the square a = 1 is required.
c = (b/2)^2
the square term is b/2
x^2 + 26x + c
a = 1 so we are okay to complete the square.
b = 26
c = (26/2)^2 = 13^2 = 169
finally to complete the square:
x^2 + 26x + 169 = (x+b/2)^2 = (x+13)^2
6 0
3 years ago
A dilation has a center​ (0,0). Find the image of point B (7/5,-5,4) for the scale factor 1/6
AveGali [126]

Answer:

Multiply the coordinate by the scale factor, 1/5 to get (7/25,-5/20).  You can simplify this to (7/25, -1/4)

5 0
3 years ago
Ammonia at 70 F with a quality of 50% and a total mass of 4.5 lbm is in a rigid tank with an outlet valve at the bottom. How muc
Inessa05 [86]

Answer:

0.10865 killograms

Step-by-step explanation:

calculating the liquid mass of ammonia removed through the bottom value from a rigid tank at constant temperature.

Given:

temperature: T=70 F

quality : 50% = 0.5

initial mass: m1= 4.5 lbm

to find the removed liquid mass first we have to find total volume from which we can find remaining mass. as the tang is rigid the temperature and volume remains constant.

by taking the difference of mass we can determine the mass of liquid removed.

 we have two phases at temperature T= 70 F with specific volume for liquid  vf=0.02631 ft^3/lbm  and specific volume for vapor is  vg=2.3098 ft^3/lbm .

The Volume in the initial state is given by, (Using definition of specific volume)

                                          V= m1v1

 using v1=x(vf+vg)

                                    V= m1x(vf+vg)

substituting m1= 4.5 lbm\\ , vf= 0.02631 ft^3/lbm , vg=2.3098 ft^3/lbm

we get

        V= (4.5 lbm)(0.5)(0.02631 ft^3/lbm +2.3098 ft^3/lbm)  

finally          V=5.2625 ft^3  

we know the formula to find liquid mass is

mass =density *volume

density of ammonia is  0.73 kg/m^3

inserting the values into the formula we get the value for liquid mass removed through the valve.

m = (0.73 kg/m^3)*(5.25625 ft^3)

the final answer is

                           m= 0.10865 kg

4 0
4 years ago
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