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zlopas [31]
3 years ago
15

What is the perimeter of a rectangle with a length of 4x-6 and a width of 2x+3

Mathematics
2 answers:
Katarina [22]3 years ago
8 0
12x - 6

I showed my work in the image attached.

Leona [35]3 years ago
4 0

Answer:

I got p=12 i'm so sorry if the answer is wrong

Step-by-step explanation:

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Suppose a white dwarf star has a diameter of approximately 1.8083 to the power of 4 km. Use the formula 4n to the power of 2 to
aivan3 [116]

ANSWER:

The surface area of the star is 3.2700 x 10^{8} square kilometres.

EXPLANATIONS:

Diameter of the star = 1.8083 x 10^{4} Km.

Surface area of the star = 4n^{2}

Where n is the radius of the star.

So that;

n = \frac{ 1.8083 * 10^{4} }{2}

   = 0.90415 x 10^{4}

n =  0.90415 x 10^{4} Km

Thus,

Surface area = 4 x (0.90415*10^{4} )^{2}

                     = 326994889

Surface area = 3.2700 x 10^{8} km^{2}

Therefore, the surface area of the star is 3.2700 x 10^{8} square kilometres.

4 0
2 years ago
What is the meaning of the unknown factor or quotient
Cloud [144]

The quotient is the answer to a division problem.
3 0
3 years ago
Given line m is parallel to line n. What theorem or postulate justifies the statement? ∠1 ≅ ∠4
natima [27]

Answer:

corresponding angles postulate

Step-by-step explanation:

just took the test ;)

6 0
3 years ago
A=1/2(a+b)h solve for h
max2010maxim [7]
H= 2a/a+b woolud be the answer
5 0
2 years ago
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.5 parts
DochEvi [55]

Answer:

We accept  H₀   with the information we have, we can say level of ozone is under the major limit

Step-by-step explanation:

Normal Distribution

population mean  =   μ₀   =  7.5   ppm

Sample size     n   =  16      df  =  n  -  1  df  =  15

Sample mean      =    μ   =  7.8  ppm

Sample standard deviation   =  s  =  0.8

We want to find out if ozono level, is above normal level that is bigger than 7.5

1.- Hypothesis Test

null hypothesis                          H₀         μ₀   =  7.5

alternative hypothesis             Hₐ          μ₀  >  7.5  

2.-Significance level    α  =  0.01   we will develop one tail-test (right)

then   for   df  =  15    and    α =  0,01   from t -student table we get

t(c)  = 2.624

3.-Compute  t(s)

t(s)  =  (  μ -  μ₀ ) / s /√n            ⇒   t(s)  = ( 7.8  -  7.5 )*4/0.8

t(s)  = 0.3*4/0.8

t(s)  = 1.5

4.-Compare   t(s)   and  t(c)

t(s)  <  t(c)       1.5  <  2.64

Then  t(s) is inside the acceptance region. We accept  H₀

7 0
2 years ago
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