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3 years ago

Solve 3(x - 2) < 18.

Mathematics

1 answer:

Anni [7]3 years ago

3 0

- {x | x < 8}

- Divide each term by 3 and simplify.

- x - 2 < 6

- Move all terms not containing x to the right side of the inequality.

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PLS HELP THIS IS A TEST

Pie

Answer:

D: 7

Step-by-step explanation:

1.75 (7) + 4.50

12.25 + 4.50

= 16.75

16.75 < 20

8 0

3 years ago

Solve the system of linear equations for x and y.(cos θ)x + (sin θ)y=1(−sin θ)x + (cos θ)y = 0x=y=

lys-0071 [83]

Answer:

$\bold{x =cos\theta}\\\bold{y=sin\theta}$

Step-by-step explanation:

The given system of linear equations is:

$(cos\theta)x+(sin\theta)y=1\\(-sin\theta)x+(cos\theta)y=0$

We have to solve the equations for the values of $x, y$ .

Let us use elimination method in which we eliminate one of the variables from the two variables.

For this, let us multiply the first equation by $sin\theta$ and second equation by $cos\theta$

Now, the equations become:

$(cos\theta.sin\theta)x+(sin\theta.sin\theta)y=sin\theta\\\Rightarrow (cos\theta.sin\theta)x+(sin^2\theta)y=sin\theta ....... (1)\\\\(-sin\theta.cos\theta)x+(cos\theta.cos\theta)y=0\\\Rightarrow (-sin\theta.cos\theta)x+(cos^2\theta)y=0 ..... (2)$

Now, let us add (1) and (2):

$(sin^2\theta)y+(cos^2\theta)y=sin\theta\\\Rightarrow (sin^2\theta+cos^2\theta)y=sin\theta\\\Rightarrow (1)y=sin\theta\\\Rightarrow y = sin\theta$

Using the equation:

$(-sin\theta)x+(cos\theta)y=0$

Putting value of $y$ :

$\Rightarrow (-sin\theta)x+(cos\theta)sin\theta=0\\\Rightarrow (sin\theta)x=(cos\theta)sin\theta\\\Rightarrow x = cos\theta$

So, the answer to the system of linear equations is:

$\bold{x =cos\theta}\\\bold{y=sin\theta}$

7 0

2 years ago

There is a line that includes the point (6,9) and has a slope of 3. What is its equation in

Nostrana [21]

Answer:

y=3x-9

Step-by-step explanation:

6 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

Read 2 more answers

Please help stuck on question

vazorg [7]

Answer:

Suppose we have a polynomial of degree N with a leading coefficient A and roots {x₁, x₂, ..., xₙ}

We can write this polynomial as:

P(x) = A*(x - x₁)*(x - x₂)*...*(x - xₙ)

such that the terms:

(x - x₁), (x - x₂), etc...

are called the factors.

In this case, we know that the roots OF THE FACTORS

are:

(x = - 2)

(x = - (1 + √5))

(x = + 3i)

If the root of the polynomial is x = -2, then the factor should be:

(x + 2)

which is zero when we evaluate x in -2

Then the correct option is the first one.

5 0

2 years ago