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In-s [12.5K]
3 years ago
15

A car rental company offers two plans for renting a car. Plan A: $35 per day and $O.25 per mile. Plan B: $55 per day with free u

nlimited mileage. For what range of miles will plan B save you money? A). Less than 150 mi. B). More than 80 mi. C). Less than 80 mi. D). Less that 400 mi. E). More than 400 mi.
Mathematics
1 answer:
salantis [7]3 years ago
4 0
The answer for this question is 90.25$
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Three girls search for geodes in the desert. corinne collected 11 1/8 pounds, ellen collected 4 5/8 pounds, and leonda collected
jeyben [28]
9 3/4. Is the answer
4 0
2 years ago
Geometry, any idea?
Pepsi [2]

Answer:

8

Step-by-step explanation:

15x-31+9×+11+×=180

25×-20=180

25×=200

×=8

15(8)-31=89

9(8)+11=83

8

5 0
3 years ago
4. Entomologists introduce 20 of one
Lina20 [59]

Answer:

This number can be found using the Rule of 70, which describes the formula dt = 70 / r, where dt is the doubling time, and r is the annual rate of growth.

Step-by-step explanation:

24/6 = 4

20 x 2 = 40 x 2 = 80 x 2 = 160 x 2 = 320

20 ---> 320 in the first day.

Keep doing this equation. Multiply each day's end result by 2 four times for 9 more days.

8 0
3 years ago
Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$
Anna [14]

Answer:

\large\boxed{1\dfrac{1}{3}\ u^2}

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}

6 0
3 years ago
A company makes traffic signs one of those signs can be modeled by an equilateral triangle with the perimeter of 144 inches the
barxatty [35]

Answer:

The height of the larger traffic signs is 51.96 inches

Step-by-step explanation:

Given that initially, A company makes traffic signs of an equilateral triangle with the perimeter of 144 inches

Now, A company makes similar traffic signs of perimeter 1.25 of original

New perimeter will be 1.25x144=180 inches

Let triangle ABC be a larger triangle with 180inches of the perimeter.

It is given that the triangle is an equilateral triangle,

Therefore, Side AB=BC=AC

The perimeter of a triangle ABC is

AB+BC+AC=180

3AB=180

AB=60 inches

Figure show that triangle ABC and AD is the height of the triangle.

Since, the triangle ABC is an equilateral triangle,

AD is the perpendicular bisector of BC

so that,

BD=DC and Angle ADB=90

In right triangle ADB,

Angle D =90 and Angle B =60

Side BD=60/2=30 inches and AB=60 inches

Using Pythagoras theorem,

AD^{2} +BD^{2} =AC^{2}

(AD)^{2} +(30)^{2} =60^{2}

(AD)^{2} =60^{2}-(30)^{2}

AD=51.96 inches

Thus, The height of the larger traffic signs is 51.96 inches

4 0
3 years ago
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