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3 years ago

Please help. I really need help with this so please help.

Mathematics

1 answer:

Flauer [41]3 years ago

4 0

Answer:

5x ≥ 1y

Step-by-step explanation:

The number key chains is 8 times the number of thumb drives. She wants to use at least 5 times at many.

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3 years ago

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Answer:

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Step-by-step explanation:

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4 years ago

Find the area of the shape.

tatiyna

Answer:

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3 years ago

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Find the vertex and the axis of symmetry of the graph of y=-6(x+4)^2 – 3.

vesna_86 [32]

Answer:

vertex = (- 4, - 3 ), equation is x = - 4

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

y = 6(x + 4)² - 3 ← is in vertex form

with (h, k ) = (- 4, 3 ) ← vertex

The equation of the axis of symmetry is a vertical line passing through the vertex, with equation

x = - 4

4 0

3 years ago

Find a formula for Y(t) with Y(0)=1 and draw its graph. What is Y\infty?

tino4ka555 [31]

Answer:

$(a)\ y(t)\ =\ -2e^{-2t}+3$

$(b)\ y(t)\ =\ 4e^{-2t}-3$

Step-by-step explanation:

(a) Given differential equation is

Y'+2Y=6

=>(D+2)y = 6

To find the complementary function, we will write

D+2=0

=> D = -2

So, the complementary function can be given by

$y_c(t)\ =\ C.e^{-2t}$

To find the particular integral, we will write

$y_p(t)\ =\ \dfrac{6}{D+2}$

$=\ \dfrac{6.e^{0.t}}{D+2}$

$=\ \dfrac{6}{0+2}$

= 3

so, the total solution can be given by

$y_(t)\ =\ C.F+P.I$

$=\ C.e^{-2t}\ +\ 3$

$y_(0)=C.e^{-2.0}\ +\ 3$

but according to question

1 = C +3

=> C = -2

So, the complete solution can be given by

$y_(t)\ =\ -2.e^{-2.t}\ +\ 3$

(b) Given differential equation is

Y'+2Y=-6

=>(D+2)y = -6

To find the complementary function, we will write

D+2=0

=> D = -2

So, the complementary function can be given by

$y_c(t)\ =\ C.e^{-2t}$

To find the particular integral, we will write

$y_p(t)\ =\ \dfrac{-6}{D+2}$

$=\ \dfrac{-6.e^{0.t}}{D+2}$

$=\ \dfrac{-6}{0+2}$

= -3

so, the total solution can be given by

$y_(t)\ =\ C.F+P.I$

$=\ C.e^{-2t}\ -\ 3$

$y_(0)\ =C.e^{-2.0}\ -\ 3$

but according to question

1 = C -3

=> C = 4

So, the complete solution can be given by

$y_(t)\ =\ 4.e^{-2.t}\ -3$

4 0

3 years ago