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2 years ago

Someone please help me I need help

Mathematics

1 answer:

mina [271]2 years ago

4 0

I think it would be 60 degrees

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AleksandrR [38]

im pretty sure the answer is C

5 0

2 years ago

Which of these could be the graph of F(x) = ln x + 2?

vodomira [7]

Could you past the graphs. I would like to help:)

7 0

3 years ago

Read 2 more answers

Construct a quadrilateral ABCD, AB = 4cm BC = 3cm,<br> AD= 2.5cm, AC= 4.5cm, BD= 4cm.

hammer [34]

Answer:

steps below

Step-by-step explanation:

1. use ruler to draw a segment of 4 units AB

2. center the compasses needle at A and draw arc with radius of 5 and intersect with another arc centered at B with radius of 3 at point C

3. center the compasses needle at B and draw arc with radius of 4.5 and intersect with another arc centered at A with radius of 2.5 at point D

4. connect BC, CD and AD to form quadrilateral ABCD

6 0

2 years ago

GUYS PLEASE HELP ME PLEASE HELP HELP PLEASE HELP ASAP GUYS HELP!!!!!

slega [8]

Answer:

$SA=5\ yd^2$

Step-by-step explanation:

we know that

The surface area of the square pyramid is equal to the area of the square base plus the area of its four triangular faces

$SA=b^2+4[\frac{1}{2}(b)(h)]$

we have

$b=1\ yd\\h=2\ yd$

substitute

$SA=1^2+4[\frac{1}{2}(1)(2)]$

$SA=5\ yd^2$

5 0

3 years ago

Inveres laplace transform (3s-14)/s^2-4s+8

Olenka [21]

Complete the square in the denominator.

$s^2 - 4s + 8 = (s^2 - 4s + 4) + 4 = (s-2)^2 + 4$

Rewrite the given transform as

$\dfrac{3s-14}{s^2-4s+8} = \dfrac{3(s-2) - 8}{(s-2)^2+4} = 3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}$

Now take the inverse transform:

$L^{-1}_t\left\{3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3L^{-1}_t\left\{\dfrac{s-2}{(s-2)^2+2^2}\right\} - 4L^{-1}_t\left\{\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3e^{2t} L^{-1}_t\left\{\dfrac s{s^2+2^2}\right\} - 4e^{2t} L^{-1}_t\left\{\dfrac{2}{s^2+2^2}\right\} \\\\ \boxed{3e^{2t} \cos(2t) - 4e^{2t} \sin(2t)}$

5 0

2 years ago