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zhannawk [14.2K]
3 years ago
14

A triangle has sides of 2 inches and 7 inches which is a possible lenght,in inches of the third side of the triangle

Mathematics
1 answer:
noname [10]3 years ago
4 0
I feel that you might have meant to include choices. However, the options must fall within this range. 

5<x<9

This is due to the fact that 2 sides combined must always be greater than the 3rd. 
You might be interested in
Find the product (4x-5y)^2
sleet_krkn [62]
To find the product of (4x-5y)^2,
we can rewrite the problem as:
(4x-5y)(4x-5y) (two times because it is squared)

Now, time to use that old method we learned in middle school:
FOIL. (Firsts, Outers, Inners, and Lasts)

FOIL can help us greatly in this scenario.
Let's start by multiplying the 'Firsts' together:
4x * 4x = <em>16x^2</em>

Now, lets to the 'Outers':
4x * -5y = <em>-20xy</em> 

Next, we can multiply the 'Inners':
-5y * 4x = <em>-20xy</em>

Finally, let's do the 'Lasts':
-5y * -5y = <em>25y</em>^2

Now, we can take the products of these equations from FOIL and combine like terms. We have: 16x^2, -20xy, -20xy, and 25y^2.
-20xy and -20xy make -40xy.

The final equation (product of (4x-5y)^2) is:
16x^2 - 40xy + 25y^2

Hope I helped! If any of my math is wrong, please report and let me know!
Have a good one.
4 0
3 years ago
Please answer this correctly
marshall27 [118]

Answer:

hello

Step-by-step explanation:

u sayh-e-l-l-o yW

6 0
3 years ago
Anyone know this because I need help !!!!
Mila [183]

Answer:

90 clockwise

Step-by-step explanation:

rotate it by 90 degrees. it lines up

5 0
3 years ago
Read 2 more answers
Assume that​ women's heights are normally distributed with a mean given by mu equals 62.5 in​,and a standard deviation given by
Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with  

\mu = 62.5 in

\sigma = 2.2 in

the normal probability density function is given by  

f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}, then

(a) P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx = 0.5899

   (in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking P(\bar{X} < 63) where n = 37. \bar{X} is normally distributed with mean 62.5 in and standard deviation 2.2/\sqrt{37}. So, the probability density function is given by

g(x) = \frac{1}{\sqrt{2\pi}\frac{2.2}{\sqrt{37}}}\exp{-\frac{(x-62.5)^{2}}{2(2.2/\sqrt{37})^{2}}}, and

P(\bar{X} < 63) = \int\limits_{-\infty}^{63}g(x)dx = 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.

4 0
3 years ago
What is the value of x in the equation 2(x-3)+9=3(x+1)+x?
Sergio039 [100]

Answer:

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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