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vampirchik [111]
3 years ago
8

Help meeeeeee ASAP helppppppp

Mathematics
1 answer:
agasfer [191]3 years ago
8 0

Answer:

45 × 1/3 = 15

Step-by-step explanation:

15 / (1/3) = (3×15)/1 = 45

so, from 15 / (1/3) = 45

we can then adapt the equation by multiplying both sides by 1/3

15 = 45 × (1/3)

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Al oeste de Albuquerque, Nuevo México, la carretera 40 con rumbo al este es recta y tiene una fuerte pendiente hacia la ciudad.
alexgriva [62]

Usando el concepto de pendiente, se encuentra que el cambio en la distancia horizontal es de 10 pies.

La pendiente de una reta es la <u>razón entre el cambio vertical y el cambio horizontal,</u> o sea, cual el cambio vertical cuando el cambio horizontal es de 1.

En este problema, hay que:

  • La pendiente es -100, o sea, cuando hay un desplazamiento de un pie por adelante, hay una queda de 100 pies.
  • Bajado 1000 pies, entonces, aplicando la proporción:

1 pie H -> -100 pies V

x pie H -> -1000 pies V

Aplicando la multiplicación cruzada:

-100x = -1000

100x = 1000

x = \frac{1000}{100}

x = 10

El cambio en la distancia horizontal es de 10 pies.

Un problema similar es dado en brainly.com/question/24766917

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3 years ago
Find the six trig function values of the angle 240*Show all work, do not use calculator
-BARSIC- [3]

Solution:

Given:

240^0

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

sin240^0=sin(180+60)

Using the trigonometric identity;

sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny

Hence,

\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

cos240^0=cos(180+60)

Using the trigonometric identity;

cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny

Hence,

\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\  \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

tan240^0=tan(180+60)

Using the trigonometric identity;

tan(180+x)=tan\text{ }x

Hence,

\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\  \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}

To get cosec 240 degrees:

\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\  \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\  \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}

To get sec 240 degrees:

\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\  \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\  \\ Thus, \\ sec240^0=-2 \end{gathered}

To get cot 240 degrees:

\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\  \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\  \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\  \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}

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1 year ago
Find the area of this circle. Use 3.14 to approximate Pi
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\pi {r}^{2} \\  \:  \:  \:  \:  \:  \:  = 3.14 \times 13 {cm}^{2}  \\  = 530.66 {cm}^{2}

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2 years ago
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