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Alex Ar [27]
2 years ago
12

Select all of the transformations that could change the location of the asymptotes of a cosecant of secant function. vertical sh

ift phase shift vertical stretch period change
Mathematics
2 answers:
Serjik [45]2 years ago
8 0

Answer:

B and D

Step-by-step explanation:

liubo4ka [24]2 years ago
4 0

Answer: Phase Shift, Period Change

Step-by-step explanation:

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The number. Find the nume.
aliya0001 [1]

Answer: 8x+3x=12x+9

The number is -9.

Step-by-step explanation: Combine like terms.

11x=12x+9

Subtract 12x from the other side. 11x-12x= -1x

-1x+9

Divide by -1 to isolate x.

-1/-1 cancels out.

9/-1= -9

x=-9

Plug in -9 to check your answer if you want to double check.

3 0
3 years ago
Help plz i need a anwser
Blababa [14]

Answer:

Your answer would be B: $25.20

24 x 5% is 1.2

$24 + 1.2 is 25.2

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Write each equation in slope-intercept form (3,4); and (-6,-2)
Alborosie

Answer:

y = 2/3x + 2

Step-by-step explanation:

Slope intercept form is y=mx+b where m is the slope and b is the y intercept.

Given 2 points you can find the slope using (y-y1)/(x-x1)

m = (-2-4)/(-6-3) = -6/-9 = 2/3

To find b use one of the points and m and plug into the slope intercept form and then solve for b.

y=mx+b

4=2/3(3) + b

4=2 + b

b=2

Now we can write the final equation as : (plug m and b back in)

y = 2/3x + 2

7 0
2 years ago
Read 2 more answers
Hi. Sorry this is bad quality. <br><br> What is the slope of the line on the graph?
d1i1m1o1n [39]
Slope (m) =0

θ =180°

distance (d) =15

ΔX =15

ΔY =0
5 0
2 years ago
Un submarino se encuentra 50m por debajo del nivel del mar y comienza a subir 2m cada 5 minutos ¿a que profundidad se encuentra
Trava [24]

Answer:

a) El submarino se encuentra a una profundidad de 26 metros una hora después de haber comenzado el ascenso.

b) Al submarino le falta 1 hora y 5 minutos para llegar a la superficie.

Step-by-step explanation:

a) De acuerdo con el enunciado, la cantidad de metros ascendidos por el submarino es directamente proporcional al tiempo, entonces la profundidad del submarino a la hora de haber comenzado el ascenso se obtiene mediante la siguiente relación lineal:

h = 50\,m - \frac{2\,m}{5\,min}\times 60\,min

h = 26\,m

El submarino se encuentra a una profundidad de 26 metros una hora después de haber comenzado el ascenso.

b) En primer lugar, calculamos el tiempo total del viaje mediante la siguiente relación:

\frac{2\,m}{5\,min} = \frac{50\,m}{t}

t = 50\,m\times \frac{5\,min}{2\,m}

t = 125\,min (2\,h\,5\,min)

El tiempo faltante para alcanzar la superficie se calcula mediante la siguiente sustracción:

\Delta t = 125\,min - 60\,min

\Delta t = 65\,min (1\,h\,5\,min)

Al submarino le falta 1 hora y 5 minutos para llegar a la superficie.

7 0
2 years ago
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