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Sidana [21]
2 years ago
7

Vector v is shown in the graph.

Mathematics
1 answer:
Elis [28]2 years ago
6 0

Question 4

<h3>Answer: Choice D</h3>

Explanation:

If the initial point is the origin, the coordinates of the terminal point form the vector itself in component form. We go from (-8,6) to <-8,6>. The notation change is from ordered pair to vector format.

We have a right triangle with legs of 8 and 6 units. The pythagorean theorem will help us determine the hypotenuse is 10. Therefore, the vector length is 10 units and we would say ||v|| = 10. We have a 6-8-10 right triangle.

===========================================================

Question 5

<h3>Answer: Choice B</h3>

Explanation:

Vector v starts at (2,5) and ends at (-3,-2).

The x component of the vector is x2-x1 = -3-2 = -5 meaning we move 5 units to the left when going from the start point to the endpoint.

At the same time we move 7 units down because y2-y1 = -2-5 = -7 which is the y component of the vector.

The component form of vector v is

v = <-5, -7>

it says "move 5 units left, 7 units down".

Apply the pythagorean theorem to find the length of the vector.

a^2+b^2 = c^2

c = sqrt(a^2 + b^2)

||v|| = sqrt( (-5)^2 + (-7)^2 )

||v|| = 8.602 which is approximate.

Now let's use the arctangent function to find the angle

theta = arctan(b/a)

theta = arctan(-7/(-5))

theta = 54.462 which is also approximate.

There's a problem however. This angle is in Q1 but the vector <-5,-7> is in Q3. An easy fix is to add on 180 to rotate to the proper quadrant.

54.462+180 = 234.462

which is the proper approximate angle for theta.

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Which numbers have a digit in the ones place that is 1/10 the value of the digit in the tens place
liq [111]

Answer:

4,099  and 5,011

Step-by-step explanation:

This problem can be solved by taking options one by one.

Option (1) : 4,099  

Digit in ones place = 9

The value of the digit in tens place = 90

\dfrac{9}{90}=\dfrac{1}{10}. It is correct.

Option (2) : 4,110

Digit in one places = 0

The value of the digit in tens place = 10

It is incorrect.

Option (3) : 5,909

Digit in one places = 9

The value of the digit in tens place = 0

It is again incorrect.

Option (4) : 5,011

Digit in one places = 1

The value of the digit in tens place = 10

\dfrac{1}{10}. It is correct.

Hence, in option (a) and (d), the he ones place is 1/10 the value of the digit in the tens place.

7 0
3 years ago
HEEEELP PLEASE<br> I dont understand
lesya [120]

Answer:

<h2>y = 0.5x</h2>

Step-by-step explanation:

\dfrac{y}{x}=a\to y=ax\\\\\text{We have}\ x=1,\ y=0.5.\\\\\text{Substitute:}\\\\a=\dfrac{0.5}{1}=0.5\\\\\text{Therefore}\ y=0.5x

7 0
3 years ago
Read 2 more answers
¿Para cuál(es) valor(es) de p las rectas de ecuación x − 1/p = 2 − y/p y x − 1/1 − p = y − 2/2 son perpendiculares? A) Solo para
Akimi4 [234]

Respuesta:

C) Solo para el -1

Explicación paso a paso:

Para resolver este problema, debemos de determinar la pendiente en cada una de las ecuaciones provistas:

\frac{x-2}{p}=\frac{2-y}{p}

y

\frac{x-1}{1-p}=\frac{y-2}{2}

ahora bien, necesitamos conocer el valor de la pendiente de una de las dos ecuaciones. Tomemos la primera ecuación y resolvámosla para y:

\frac{x-2}{p}=\frac{2-y}{p}

Multiplicamos ambos lados para p y obtenemos:

x-1=2-y

volteamos la ecuación y nos da:

2-y=x-1

pasamos el 2 a restar al otro lado y nos da:

-y=x-1-2

-y=x-3

y dividimos ambos lados de la ecuación dentro de -1

y=-x+3

esta ecuación ya tiene la forma pendiente intercepto:

y=mx+b

donde m es nuestra pendiente:

m_{1}=-1

Esta es la pendiente de una de las dos ecuaciones, para que la segunda ecuación sea perpendicular a la primera, su pendiente debe de ser el recíproco negativo de la pendiente de la primera ecuación, entonces la pendiente de la segunda ecuación debe ser:

m_{2}=-\frac{1}{m_{1}}

m_{2}=-\frac{1}{-1}

m_{2}=1

ahora tomamos la segunda ecuación y encontramos su pendiente. Tomemos la ecuación:

\frac{x-1}{1-p}=\frac{y-2}{2}

y despejemos y, comenzamos multiplicando ambos lados de la ecuación por 2, así que obtenemos:

2\frac{x-1}{1-p}=y-2

Multiplicamos el 2 por cada término de la fracción, entonces obtenemos:

\frac{2x-2}{1-p}=y-2

ahora pasamos el 2 a sumar al lado izquierdo y obtenemos:

\frac{2x-2}{1-p}+2=y

Ahora podemos separar la fracción del lado izquierdo en dos fracciones para obtener:

\frac{2x}{1-p}-\frac{2}{1-p}+2=y

volteamos la ecuación y nos da:

y=\frac{2x}{1-p}-\frac{2}{1-p}+2

Ahora nuestra ecuación ya tiene la forma y=mx+b

de aquí podemos determinar nuestra pendiente:

m=\frac{2}{1-p}

Con la primera ecuación determinamos que esta pendiente debería de ser igual a 1, entonces igualamos esa segunda pendiente a 1 para obtener:

\frac{2}{1-p}=1

y despejamos p

Pasamos a multiplicat el 1-p al lado derecho de la ecuación para obtener:

2=1-p

volteamos la ecuación:

1-p=2

pasamos el 1 a restar al lado derecho:

-p=2-1

-p=1

y multiplicamos ambos lados de la ecuación por -1 para obtener:

p=-1

Entonces la respuesta es C) solo para el -1

4 0
3 years ago
Bonnie deposits $70.00 into a new savings account. The account earns 45 % simple interest per year , s added or removed from the
nordsb [41]

Answer:

\$164.5

Step-by-step explanation:

Bonnie deposits $70.00 into a new savings account.

The account earns 45% simple interest per year.

She neither added or removed from the savings account for 3 years.

We know that,

i=\dfrac{P\cdot r\cdot r}{100}

here,

i = interest,

P = principal = $70,

r = rate of interest = 45%,

t = time = 3 years,

Putting the values,

i=\dfrac{70\cdot 45\cdot 3}{100}=\$94.5

So the total amount will be,

=\text{Principal}+\text{Interest}

=70+94.5

=\$164.5

6 0
3 years ago
Read 2 more answers
Andre and Elena are each saving money. Andre starts with $100 in his savings
zysi [14]
Andre’s equation is y = 5x + 100 and Elena’s is y = 20x + 10
Andre - 5x + 100 the x is how many weeks (6) so it’ll be 5(6) + 100 which is 30 + 100 = $130 in 6 weeks
Elena - 20x + 10 same thing x is how many weeks (6) as well so 20(6) + 10 which is 120 + 10 = $130 as well in 6 weeks so therefore, the answer is C.
4 0
2 years ago
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