2 Points

[ Trigonometry ]

PSYCHO15rus [73]

Hello from MrBillDoesMath!

Answer:

B = 67.4 degrees (approximately)

Discussion:

From plane trigonometry

sin(B) = side opposite angle B \ hypotenuse

= 12/13

Use a calculator (take the inverse sine of 12/13) gives

B = 67.4

Thank you,

MrB

3 0

3 years ago

Albert is traveling 60 mph in his car. After 3 hours, how far will he have traveled?

Dahasolnce [82]

He will have traveled 180 miles, though with real situations the distance may vary.

6 0

3 years ago

Read 2 more answers

Which statement describes the inverse of m(x) = x2 – 17x?

stealth61 [152]

Answer:

The correct option is;

$The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}$

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

$x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

$x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}$

Which can be simplified as follows;

$x = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2} \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}$

And further simplified as follows;

$x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}$

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

$m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}$

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

$The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}$ to increase m⁻¹(x) above the minimum.

8 0

2 years ago

RSM QUESTION NEED HELP!!!!!! A pharmacist has a 13 % alcohol solution and another 18 % alcohol solution. How much of each must h

Luden [163]

Answer:

40 grams of 13% alcohol solution and 10 grams of 18% alcohol solution.

Step-by-step explanation:

13%(x) + 18%(50-x) = 14%*50g Multiply them all

0.13x + (9-0.18x) = 7

0.13x-0.18x + 9 = 7 Simplify the equation and subtract 0.13x from 0.18x

9-0.05x = 7 Add 0.05x to each side

9 = 7+0.05x Subtract 7 from both sides

0.05x = 2 Multiply each side by 100

5x = 200 Divide both sides by 5

x = 40

50 - 40 = 10

40 grams and 10 grams

Your welcome

4 0

3 years ago

Read 2 more answers

A box is being created out of a 15 inch by 10 inch sheet of metal. Equal-sized squares are cutout of the corners, then the sides

ivolga24 [154]

Answer:

Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box has largest volume.

Step-by-step explanation:

Given that,

A box is being created out of a 15 inches by 10 inches sheet of metal.

The length of the one side of the squares which are cut out of the each corners of the metal sheet be x.

The length of the metal box be = (15-2x) inches.

The width of the metal box be =(10-2x) inches

The height of the metal box be =x inches

Then, the volume of the metal box= length×width×height

=(15-2x)(10-2x)x cubic inches

=(150x-50x²+4x³) cubic inches

∴ V= 4x³-50x²+15x

Differentiating with respect to x

V'=12x²-100x+15

Again differentiating with respect to x

V''=24x-100

For maximum or minimum value, V'=0

12x²-100x+15=0

Apply quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ , here a=12, b= -100 and c=15

$x=\frac{-(-100)\pm\sqrt{(-100)^2-4.12.15}}{2.12}$

$\Rightarrow x=\frac{100\pm\sqrt{9280}}{2.12}$

$\Rightarrow x=0.1528,8.18$

For x= 8.18, The value of (15-2x) and (10-2x) will negative.

∴x=0.1528 .

Now, $V''|_{x=0.1528}=24(0.1528)-100$

∴At x=0.1528 inch, the volume of the metal box will be maximum.

Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box has largest volume.

4 0

3 years ago