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Snezhnost [94]
3 years ago
8

The quadratic x? - x = 2 has two solutions. What is the larger of the two solutions?

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
5 0
I believe the 2 is the largest
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Please help me....I really need help!!
scoundrel [369]

\dfrac{c^2-4}{6c^4+15c^3}\div\dfrac{c^2+4c+4}{12c^3+30c^2}=\dfrac{c^2-2^2}{3c^3(2c+5)}\div\dfrac{c^2+2(c)(2)+2^2}{6c^2(2c+5)}\\\\_{\text{use}\ a^2-b^2=(a-b)(a+b)\ \text{and}\ (a+b)^2=a^2+2ab+b^2}\\\\=\dfrac{(c-2)(c+2)}{3c^3(2c+5)}\cdot\dfrac{6c^2(2c+5)}{(c+2)^2}=\dfrac{(c-2)}{c}\cdot\dfrac{2}{(c+2)}=\boxed{\dfrac{2(c-2)}{c(c+2)}}

6 0
3 years ago
Solve 6 + 5 √ 2 4 9 − 2 x = 7
NISA [10]

6+5\sqrt{249}-2x=7 \\-2x=7-6-5\sqrt{249} \\-2x\approx-77.9 \\x\approx\frac{-77.9}{2}\approx38.95

Hope this helps.

7 0
3 years ago
if you could respond with the work shown that would be amazing! i rate people very nicely!! this is due before 7 AM EST time :)
yulyashka [42]

Answer:

35ft squared

Step-by-step explanation:

First lets try solving the rectangular middle part

Width is 3 ft

Height is 5ft + 4ft = 9ft

3*9ft= 27ft. This is area of middle rectangular section

For the triangle, we subtract 3 from the rectangle from 11ft and divide by 2 because the two triangle on the edges are the same.

11-3=8

8/2=4

Triangle area formula = Base*Height / 2

4*4/2

16/2

4ft squared

and the other triangles area is also 4ft squared because they are congruent. So you add all the sections of the area up

27ft+4ft+4ft=35ft squared

35ft squared is the answer.

Hope this helps :)

4 0
3 years ago
3x~6y=~3<br> 2x+4y=30<br> Elimination using multiplication
Nat2105 [25]
\left \{ {{3x-6y=-3\ \ |*2} \atop {2x+4y=30\ \ | *(-3)}} \right. \\\\ \left \{ {{6x-12y=-6} \atop {-6x-12y=-90}} \right. \\+----\\elimination\\\\&#10;0 \neq -69\\\\There\ are\ no\ solutions
7 0
3 years ago
An equilateral triangle has a side length of 6. What is the height of the triangle?
vlabodo [156]
Since the height of an equilateral triangle in terms of its side s is s√3/2, the height of the triangle is 6√3/2 = 3√3 and so the area is (1/2)(6)(3√3) = 9√3. 

<span>If we draw a horizontal line a height of h from the base of the triangle, the region is split into two regions: the lower region consisting of a trapezoid of height h and the upper region consisting of a triangle of height 3√3 - h. </span>

<span>Since the upper triangle and the triangle itself are similar triangles, the base and height are proportional. If we let x denote the base of the length of the upper triangle, we have: </span>

<span>(S. of small triangle)/(S. of big triangle) = (Ht. of small triangle)/(Ht. of big triangle) </span>
<span>==> x/6 = (3√3 - h)/(3√3) </span>
<span>==> x = (6√3 - 2h)/√3 </span>

<span>Thus, the area of the upper triangle is: </span>

<span>A = (1/2)[(6√3 - 2h)/√3](3√3 - h) = [(6√3 - 2h)(3√3 - h)]/(2√3). </span>
<span>(Made a dumb mistake about the height here for some reason) </span>

<span>Since we require that the area of this triangle is to be half of the total area (9√3/2), we need to solve: </span>

<span>[(6√3 - 2h)(3√3 - h)]/(2√3) = 9√3/2 </span>
<span>==> (6√3 - 2h)(3√3 - h) = 27 </span>
<span>==> 54 - 6h√3 - 6h√3 + 2h^2 = 27 </span>
<span>==> 2h^2 - 12h√3 + 27 = 0. </span>

<span>Solving with the Quadratic Formula gives: </span>

<span>h = (6√3 + 3√6)/2 ≈ 8.87 units and h = (6√3 - 3√6)/2 ≈ 1.52 units. </span>

<span>Since h = (6√3 + 3√6)/2 would place the line outside of the triangle, we pick h = (6√3 - 3√6)/2. </span>

<span>Therefore, the line should be ==> (6√3 - 3√6)/2 units from the base. </span>

<span>I hope this helps! ^^ Brainliest Please?</span><span>
</span>
5 0
3 years ago
Read 2 more answers
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