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3 years ago

Geometry Help:

Mathematics

1 answer:

notka56 [123]3 years ago

6 0

1) E&B and K&A

2) X&L and E&E

3) K&L and A&F

4) C&L and I&E

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Solve : <br>4² ( 4³ )<br>with work shown

stepan [7]

Answer:

1024

Step-by-step explanation:

4²(4³)

4×4(4×4×4)

16(16×4)

16(64)

16×64

=1024

7 0

3 years ago

PLEASE HELP

kifflom [539]

$\\ \sf\longmapsto \dfrac{x^2-1}{x^2+5x+4}\leqslant 0$

Solving denominator

$\\ \sf\longmapsto x^2+5x+4>0$

$\\ \sf\longmapsto x^2+4x+x+4>0$

$\\ \sf\longmapsto x(x+4)+1(x+4)>0$

$\\ \sf\longmapsto (x+4)(x+1)>0$

$\\ \sf\longmapsto x>-4\:or x>-1$

Hence x $\neq$ -1
x can't be 0 as it makes function undefined

$\\ \sf\longmapsto -4$

5 0

3 years ago

A bulb can either be on or off. A board contains 20 bulbs connected to a randomization circuit that lights up a random sequence

Step2247 [10]

The chance that any given bulb is on is equal to 1/2. The chance that all twenty bulbs are on at the same time is $(\frac{1}{2}) ^{20} = \frac{1^{20}}{2^{20}} = \frac{1}{1048576}$

8 0

3 years ago

Read 2 more answers

Find the equation of the line that is parallel to the given line and passes through the given point. y=0,5x-2;(-2,1)

Mashcka [7]

Answer:

y = 0.5x

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 0.5x - 2 ← is in slope- intercept form

with slope m = 0.5

Parallel lines have equal slopes, thus

y = 0.5x + c ← is the partial equation

To find c substitute (- 2, 1) into the partial equation

- 1 = - 1 + c ⇒ - 1 + 1 = 0

y = 0.5x + 0 , that is

y = 0.5x ← equation of parallel line

3 0

3 years ago

In physics, if a moving object has a starting position at so, an initial velocity of vo, and a constant acceleration a, the

emmasim [6.3K]

Answer:

$a=\frac{2S -2v_ot-2s_o}{t^2}$

Step-by-step explanation:

We have the equation of the position of the object

$S = \frac{1}{2}at ^2 + v_ot+s_o$

We need to solve the equation for the variable a

$S = \frac{1}{2}at ^2 + v_ot+s_o$

Subtract $s_0$ and $v_0t$ on both sides of the equality

$S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o$

$S -v_ot-s_o = \frac{1}{2}at ^2$

multiply by 2 on both sides of equality

$2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2$

$2S -2v_ot-2s_o =at ^2$

Divide between $t ^ 2$ on both sides of the equation

$\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}$

Finally

$a=\frac{2S -2v_ot-2s_o}{t^2}$

5 0

3 years ago