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2 years ago

12

Solve the inequality n - 10 > 6.

1 answer:

givi [52]2 years ago

4 0

Answer:

$\large\boxed{n>16\to n\in(16,\ \infty)}$

Step-by-step explanation:

n - 10 > 6 <em>add 10 to both sides</em>

n - 10 + 10 > 6 + 10

n > 16

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docker41 [41]

The independent variable is:
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3 years ago

Please solve. Evaluate the function.

Yuliya22 [10]

Answer:

(g o f)(x) = $\sqrt{-510}$ (not simplified)

Step-by-step explanation:

(g o f)(x) = g(f(x))

g(x) = $\sqrt{3x}$ f(x) = $x^{3} + 9x$

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3 years ago

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anyanavicka [17]

Answer:

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Step-by-step explanation:

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2 years ago

The coordinates of the vertices of a rectangle are (−3, 4) , (7, 2) , (6, −3) , and (−4, −1) . What is the perimeter of the rect

ki77a [65]

Answrer

Find out the what is the perimeter of the rectangle .

To prove

Now as shown in the figure.

Name the coordinates as.

A(−3, 4) ,B (7, 2) , C(6, −3) , and D(−4, −1) .

In rectangle opposite sides are equal.

Thus

AB = DC

AD = BC

Formula

$Disatnce\ formula = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2}}$

Now the points A(−3, 4) and B(7, 2)

$AB = \sqrt{(7- (-3))^{2} +(2- 4)^{2}}$

$AB = \sqrt{(10)^{2} +(-2)^{2}}$

$AB = \sqrt{100+4}$

$AB = \sqrt{104}$

$AB = 2\sqrt{26}\units$

Thus

$CD= 2\sqrt{26}\units$

Now the points

A (−3, 4) , D (−4, −1)

$AD = \sqrt{(-4 - (-3))^{2} +(-1- 4)^{2}}$

$AD = \sqrt{(-1)^{2} +(-5)^{2}}$

$AD = \sqrt{1 + 25}$

$AD = \sqrt{26}\units$

Thus

$BC = \sqrt{26}\units$

Formula

Perimeter of rectangle = 2 (Length + Breadth)

Here

$Length = 2\sqrt{26}\ units$

$Breadth = \sqrt{26}\ units$

$Perimeter\ of\ rectangle = 2(2\sqrt{26} +\sqrt{26})$

$Perimeter\ of\ rectangle = 2(3\sqrt{26})$

$Perimeter\ of\ rectangle = 6\sqrt{26}$

$\sqrt{26} = 5.1 (Approx)$

$Perimeter\ of\ rectangle = 6\times 5.1$

Perimeter of a rectangle = 30.6 units.

Therefore the perimeter of a rectangle is 30.6 units.

8 0

2 years ago

Find values of  that satisfy the equation for 0º    360º and 0  θ  2 . Give answers in degrees and radians.

suter [353]

Answer:

$\theta = 30^\circ, 330^\circ$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

Step-by-step explanation:

Given [Missing from the question]

Equation:

$cos\theta = \frac{\sqrt 3}{2}$

Interval:

$0 \le \theta \le 360$

$0 \le \theta \le 2\pi$

Required

Determine the values of $\theta$

The given expression:

$cos\theta = \frac{\sqrt 3}{2}$

... shows that the value of $\theta$ is positive

The cosine of an angle has positive values in the first and the fourth quadrants.

So, we have:

$cos\theta = \frac{\sqrt 3}{2}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{\sqrt 3}{2})$

$\theta = 30$ --- In the first quadrant

In the fourth quadrant, the value is:

$\theta = 360 -30$

$\theta = 330$

So, the values of $\theta$ in degrees are:

$\theta = 30^\circ, 330^\circ$

Convert to radians (Multiply both angles by $\pi/180$ )

So, we have:

$\theta = \frac{30 * \pi}{180}, \frac{330 * \pi}{180}$

$\theta = \frac{\pi}{6}, \frac{33 * \pi}{18}$

$\theta = \frac{\pi}{6}, \frac{11 * \pi}{6}$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

8 0

2 years ago