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3 years ago

Where does the section of the bridge meet ground level? Solve 0 = -x2 + 10x - 8

Mathematics

2 answers:

scZoUnD [109]3 years ago

6 0

Answer:

The answer is (x-5)^2=17

Step-by-step explanation:

djverab [1.8K]3 years ago

4 0

In can be less mental effort if the equation is multiplied by -1 first.
.. x^2 -10x +8 = 0
.. x^2 -10x +25 +8 -25 = 0
.. (x -5)^2 = 17
.. x -5 = ±√17
.. x = 5 ±√17

The values of x that satisfy the given equation are
.. x = 5 -√17 ≈ 0.877
.. x = 5 +√17 ≈ 9.123

You might be interested in

Evaluate 3xy with an exponent of 2 y, if x=2 and y=5

Bogdan [553]

Answer:

145

Step-by-step explanation:

x = 2 , y= 5

Putting the values of x and y in the expression

=3(2) (5)^2-5

=3(2)(25)-5

=150 -5

=145

8 0

2 years ago

Help and explain thanks!

katen-ka-za [31]

Answer:

120 cm

Step-by-step explanation:

One way to tackle this is by getting another sheet of paper and drawing it out, then counting up the total of the sides. If you draw it, you can see that you're dealing with a rectangle; two sides of length 12 and two sides of length 8. If you don't like drawing or don't want to in this case, another way to get the answer is by knowing one vertex is at (0, 0), so the next vertex (0, 8), would create a side that's exactly 8 units long. Kind of the same, you know from (0, 0), you also have a point (12, 0), so drawing that would create a side that's 12 units long. All in all, to get the perimeter in units, you have 12 + 12 + 8 + 8 = 40.

The problem says it wants the amount of wood in centimeters needed for the perimeter. What we just found was the perimeter in generic units, so if the problem says every "grid square", or unit, is 3 centimeters long, then all you have to do is take our result 40 and multiply it by 3 to get the number of centimeters. Your perimeter in centimeters would be 120 cm.

7 0

2 years ago

Can anybody help me???

Tasya [4]

#10

$\\ \rm\longmapsto 0$

#11

Last three i.e Option B,C,D

#13

It started from negative ends at positive hence

$\\ \rm\longmapsto y=-|x|+2$

#14

(-3,-10)

$\\ \rm\longmapsto 2(-3)-(-10)=4$

$\\ \rm\longmapsto -6+10=4$

Proved

#15

(1,1)

$\\ \rm\longmapsto 1=(1-2)^2=(-1)^2=1$

#16

(2,-4)

$\\ \rm\longmapsto -2-2(-4)>3$

$\\ \rm\longmapsto -2+8>3$

$\\ \rm\longmapsto 6>3$

5 0

2 years ago

If hot and cold water taps are turned on altogether, a bath is filled in three minutes but if the cold tap was only half opened

Anvisha [2.4K]

9514 1404 393

Answer:

12 minutes

Step-by-step explanation:

Let c and h represent the filling times for the cold and hot taps, respectively. When the cold tap is 1/2 open, we presume that means the filling time becomes 2c.

In terms of baths per minute, the relationships are ...

1/c + 1/h = 1/3

1/(2c) +1/h = 1/(3 +1.8) . . . . . 1:48 min:sec is 1.8 minutes

Subtracting the first equation from twice the second, we get ...

2(1/(2c) +1/h) -(1/c +1/h) = 2(1/4.8) -(1/3)

1/h = 2/4.8 -1/3 = 1/12

h = 12

It takes the hot tap 12 minutes to fill the tub alone.

3 0

2 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

VladimirAG [237]

First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:

$y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}$

And the derivative of x:

$x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}$

Now, we can calculate the area of the surface:

$A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)$

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

$(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\$

$=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\$

$=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}$

Calculate indefinite integral:

$\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}$

And the area:

$A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}$

Answer D.

6 0

3 years ago

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