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3 years ago

Mr. Dewalt suggested contacting a travel agent to see whether they could get a lower price. The agent found

1 answer:

7 0

Finding a price that is 30% lower than $2075 is the same as finding a price that is 70% of $2075.

That means all we need to do is calculate what is 70% of $2075.

Convert 70% into decimal form which is 0.70.

Next, multiply $2075 by 0.70.

$2075*0.70 = $1452.50

The price the travel agent found was $1452.50

Note: If you don't know how to multiply by decimals you can multiply $2075 by 7 than divide the result by 10 (move the decimal point one place to the left).

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Last one ifr today ty !

timama [110]

Answer:

The answer is D. -3, 6, 11

Step-by-step explanation:

A coefficient is a number used to multiply a variable.

Example: 6z means 6 times z, and "z" is a variable, so 6 is a coefficient. Variables with no number have a coefficient of 1.

Example: x is really 1x.

4 0

4 years ago

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dolphi86 [110]

AB and AB’ are congruent
BC and B’C are congruent
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7 0

3 years ago

What is 3/5 * 2/3 in simplest form

Flauer [41]

Answer:

2/5

Step-by-step explanation:

3/5 * 2/3

Rewriting

3/3 * 2/5

The 3's cancel

1 * 2/5

2/5

5 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!! I NEED TO FINISH THESE QUESTIONS BEFORE MIDNIGHT TONIGHT.

hoa [83]

Answer:

The answer to your question is AC = 9.9

Step-by-step explanation:

To solve this problem use trigonometric functions. The trigonometric function that relates the hypotenuse and the adjacent side is cosine.

cos α = $\frac{adjacent side}{hypotenuse}$

Solve for hypotenuse

$hypotenuse = \frac{adjacent side}{cos\alpha }$

Substitution

hypotenuse = $\frac{9}{cos 24}$

Simplification and result

hypotenuse = 9.85

6 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

3 years ago