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matrenka [14]
3 years ago
6

Passing through (-2,8) and perpendicular to y=2x+5

Mathematics
1 answer:
Ghella [55]3 years ago
4 0
I wouldnt say its perpendicular
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Find the value of the variable.
MatroZZZ [7]

Answer:

sin 30=14/x

x=14*2

x=28

and

x=√{28²-14²}

x=14√3

7 0
3 years ago
Which of the triangles shown are right triangles?
crimeas [40]
The answer is both triangles A and D







Hope this helps :)
7 0
2 years ago
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What is the scale factor if a 4" by 10" photograph is enlarged to a poster that is 1 ft. by 2.5 ft.?
Lelechka [254]

Answer:

sorry I can't help you with this answer

7 0
3 years ago
Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1
Shkiper50 [21]

Answer:

\displaystyle  -\frac{1}{2} \leq x < 1

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0

Simplifying by x-1 and taking x=1 out of the possible solutions:

\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

(6x^3+14x^2+10x+12)

is always positive and doesn't affect the result. It can be neglected. The expression

(x-\frac{1}{2})^2

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0

For the expression to be negative, both signs must be opposite, that is

(x+\frac{1}{2})\geq 0, (x-1)

Or

(x+\frac{1}{2})\leq 0, (x-1)>0

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

\displaystyle  -\frac{1}{2} \leq x < 1

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

\boxed{\displaystyle  -\frac{1}{2} \leq x < 1 }

7 0
3 years ago
What is the answer for X and Y<br> (Geometry)
Natasha2012 [34]

Answer: y =21; x = 7√3

have: cos 60° = x/(14√3) = 1/2 => x = 1/2. (14√3) = 7√3

sin 60° = y/(14√3) = (√3)/2 => y = (√3)/2 . 14√3 = 21

Step-by-step explanation:

4 0
3 years ago
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