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3 years ago

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A man travels 25 miles to work. On his way home, he stops to fill up with gas after going d miles. Write an expression to repres

ent his distance from home.

1 answer:

lbvjy [14]3 years ago

5 0

25d or 25 + d would be the possible expressions for your question.

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Need help simple explanation if possible

photoshop1234 [79]

Answer:

m < A = 60º

m < B = 30º

Step-by-step explanation:

The given sides on this triangle are: 6, $6\sqrt{3}$ , and 12

Any triangle with the angles of 30º - 60º - 90º always has side lengths in this proportion:

x, $x\sqrt{3}$ , 2x

We can line this up with the given sides. If x is 6, then 2x would be 12.

x : $x\sqrt{3}$ : 2x = 6 : $6\sqrt{3}$ : 12

Angle B is across from 6, the shortest side. That also means that it corresponds to x, or the smallest angle in the proportion, 30º.

m < B = 30º

Solving for < A:

Method 1) Sum of Angles in a Triangle

Since we already know that one angle is right and therefore 90º and m < B is 30º, we can subtract these from the total sum of angle measures in a triangle to get the last angle, < A.

180º - 90º - 30º = 60º

m < A = 60º

Method 2) Using the second part of the proportion

Since m < A is across from the second largest side, we know that it is equal to $x\sqrt{3}$ ( $6\sqrt{3}$ in this question) or 60º in the angle proportion.

This means that m < A = 60º

Let me know if you have any questions!

4 0

2 years ago

Read 2 more answers

1. A boat is 18 ft away from a point perpendicular to the shoreline. A person stands at a point down the shoreline so that a 68°

slega [8]

Answer:

19.4 feet

Step-by-step explanation:

When calculating distances using perpendicular angles, we use trig functions such as Sin, Cos, and Tan.

Sine is defined in a perpendicular triangle as the ratio of the opposite side to the angle over the hypotenuse.
Cosine is defined in a perpendicular triangle as the ratio of the adjacent side to the angle over the hypotenuse.
Tangent is defined in a perpendicular triangle as the ratio of the opposite side over the adjacent side.

Because our angle and know side value are across from each other and we need to know the hypotenuse, we chose to use Sine.

We set up the equation $sin (68) = \frac{18}{h}$ .

We isolate h by multiplying h across the equal sign and dividing sin (68) across as well. $sin (68) = \frac{18}{h}\\hsin (68) = \frac{18}{h}*h\\hsin (68) = 18\\\frac{hsin (68)}{sin (68)} =\frac{18}{sin (68)} \\$

And finally we have $h=\frac{18}{sin (68)}$ . We input to calculator.

$h=\frac{18}{0.92718} \\h=19.4 feet$

See the attached picture below showing the shore with the boat out at sea and the position of the person.

8 0

3 years ago

Every year Victoria receives $30 for her birthday plus $2 for each year of her age Lacey receives $20 for her birthday and $4 fo

sergeinik [125]

2018 they will make the same

3 0

2 years ago

Read 2 more answers

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .

The area (in square inches) of the circle that the hexagon did not cover would be:

$\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}$ .

3 0

3 years ago