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3 years ago

An object consisting of two endpoints and all of the points between them

Mathematics

1 answer:

Over [174]3 years ago

3 0

that'd be a segment.

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Fiona spent $13.70 to travel across the city in a taxi. It cost her $3.25 to get the taxi and $2.75 each mile she traveled. Writ

Goryan [66]

$2.75m+3.25=13.70$

$2.75m=13.70-3.25\\\\2.75m=10.45\ \ \ /:2.75\\\\m=\frac{10.45}{2.75}\\\\m=\frac{1045}{275}\\\\m=3.8\ (miles)$

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3 years ago

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Pedestrian or bicycle $0.50

Zielflug [23.3K]

2 groups if 4 bicyclists or 4 groups of 2 bicyclists. Well it's only 1.75 per truck so since there were 4 trucks 4•1.75=7.00 :)

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3 years ago

Negative 22 minus 16

Maru [420]

The answer is -38.

$-22-16=-38$

4 0

3 years ago

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Please Help me with the problem please!

evablogger [386]

Answer:

35%

total number of 11th graders = 32+18 = 50

calculating percentage:

18÷50×100

= 36%

4 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