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3 years ago

10

HELP ME PLEASE!!!!!!!!!!

1 answer:

pashok25 [27]3 years ago

7 0

(7 in)^3 = (7^3) in^3 = 343 in^3

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Please help me with Alegbra 2!! Solve x2 + 4x + 1 = 0.

Julli [10]

$x^2+4x+1=0\\\\\underbrace{x^2+2x\cdot2+2^2}_{(x+2)^2}-2^2+1=0\\\\use\ (a+b)^2=a^2+2ab+b^2\\\\(x+2)^2-4+1=0\\\\(x+2)^2-3=0\\\\(x+2)^2=3\iff x+2=\sqrt3\ or\ x+2=-\sqrt3\\\\\boxed{x=\sqrt3-2\ or\ x=-2-\sqrt3}$

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3 years ago

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Barbara drives between Miami, Florida and West Palm Beach, Florida. she drives 45 miles in clear weather and then encounters a t

WINSTONCH [101]

Answer:

Barbara's speed in clear weather is $60mph$ and in the thunderstorm is $38mph$ .

Step-by-step explanation:

Let $v_1$ be the speed and $t_1$ be the time Barbara drives in clear weather, and let $v_2$ be the speed and $t_2$ be the time she drives in the thunderstorm.

Barbara drives 22 mph lower in the thunderstorm than in the clear weather; therefore,

(1). $v_2 = v_1 -22$

Also,

(2). $v_1t_1 = 45miles$

(3). $v_2 t_2 = 57$ ,

and

(4). $t_1+t_2 = 2.25hr$

From equations (2) and (3) we get:

$t_1 = \dfrac{45}{v_1}$

$t_2 =\dfrac{57}{v_2}$

putting these in equation (4) we get:

$\dfrac{45}{v_1}+\dfrac{57}{v_2}=2.25$

and substituting for $v_2$ from equation (1) we get:

$\dfrac{45}{v_1}+\dfrac{57}{v_1-22}=2.25$

This equation can be rewritten as

$2.25v_1^2-151.5v_1+990=0$

which has solutions

$v_1 = 60$

$v_1 = 7.33$

We take the first solution $v_1 =60$ because it gives a positive value for $v_2:$

$v_2 = v_1 -22$

$v_2 = 60 -22\\$

$v_2 = 38$ .

Thus, Barbara's speed in clear weather is $60mph$ and in the thunderstorm is $38mph$ .

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IceJOKER [234]

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Alenkasestr [34]

Answer:

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Step-by-step explanation:

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