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3 years ago

Which of the following is a solution of x2 + 4x + 10?

Mathematics

2 answers:

Sloan [31]3 years ago

4 0

Answer:

$x=2+-i \sqrt{6}$

Step-by-step explanation:

$x^2 + 4x + 10$

To find out the solution we set the expression =0 and solve for x

$x^2 + 4x + 10=0$

Apply quadratic formula to solve for x

$x=\frac{-b+-\sqrt{b^2-4ac}}{2a}$

a=1, b=4, c=10 plug in the values in the formula

$x=\frac{-4+-\sqrt{4^2-4(1)(10)}}{2a}$

$x=\frac{-4+-\sqrt{-24}}{2(1)}$

The value of square root (-1) is 'i'

$x=\frac{-4+-2i\sqrt{6}}{2}$

Divide each term by 2

$x=2+-i\sqrt{6}$

sattari [20]3 years ago

3 0

X^2+4x+10=0

x^2+4x=-10

x^2+4x+4=-6

(x+2)^2=-6

x+2=±i√6

x=-2±i√6

So the correct answer is the third one down from the top.

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PLEASE HELP A flight across the US takes longer east to west then it does west to east. This is due to the plane having a headwi

il63 [147K]

To solve our questions, we are going to use the kinematic equation for distance: $d=vt$
where
$d$ is distance
$v$ is speed
$t$ is time

1. Let $v_{w}$ be the speed of the wind, $t_{w}$ be time of the westward trip, and $t_{e}$ the time of the eastward trip. We know from our problem that the distance between the cities is 2,400 miles, so $d=2400$ . We also know that the speed of the plane is 450 mi/hr, so $v=450$ . Now we can use our equation the relate the unknown quantities with the quantities that we know:

<span>Going westward:
The plane is flying against the wind, so we need to subtract the speed of the wind form the speed of the plane:
</span> $d=vt$
$2400=(450-v_{w})t_{w}$

Going eastward:
The plane is flying with the wind, so we need to add the speed of the wind to the speed of the plane:
$d=vt$
$2400=(450+v_{w})t_{e}$

We can conclude that you should complete the chart as follows:
Going westward -Distance: 2400 Rate: $450-v_w$ Time: $t_w$
Going eastward -Distance: 2400 Rate: $450+v_w$ Time: $t_e$

2. Notice that we already have to equations:
Going westward: $2400=(450-v_{w})t_{w}$ equation(1)
Going eastward: $2400=(450+v_{w})t_{e}$ equation (2)

Let $t_{t}$ be the time of the round trip. We know from our problem that the round trip takes 11 hours, so $t_{t}=11$ , but we also know that the time round trip is the time of the westward trip plus the time of the eastward trip, so $t_{t}=t_w+t_e$ . Using this equation we can express $t_w$ in terms of $t_e$ :
$t_{t}=t_w+t_e$
$11=t_w+t_e$ equation
$t_w=11-t_e$ equation (3)
Now, we can replace equation (3) in equation (1) to create a system of equations with two unknowns:
$2400=(450-v_{w})t_{w}$
$2400=(450-v_{w})(11-t_e)$

We can conclude that the system of equations that represent the situation if the round trip takes 11 hours is:
$2400=(450-v_{w})(11-t_e)$ equation (1)
$2400=(450+v_{w})t_{e}$ equation (2)

3. Lets solve our system of equations to find the speed of the wind:
$2400=(450-v_{w})(11-t_e)$ equation (1)
$2400=(450+v_{w})t_{e}$ equation (2)

Step 1. Solve for $t_{e}$ in equation (2)
$2400=(450+v_{w})t_{e}$
$t_{e}= \frac{2400}{450+v_{w}}$ equation (3)

Step 2. Replace equation (3) in equation (1) and solve for $v_w$ :
$2400=(450-v_{w})(11-t_e)$
$2400=(450-v_{w})(11-\frac{2400}{450+v_{w}} )$
$2400=(450-v_{w})( \frac{4950+11v_w-2400}{450+v_{w}} )$
$2400=(450-v_{w})( \frac{255011v_w}{450+v_{w}} )$
$2400= \frac{1147500+4950v_w-2550v_w-11(v_w)^2}{450+v_{w}}$
$2400(450+v_{w})=1147500+2400v_w-11(v_w)^2$
$1080000+2400v_w=1147500+2400v_w-11(v_w)^2$
$(11v_w)^2-67500=0$
$11(v_w)^2=67500$
$(v_w)^2= \frac{67500}{11}$
$v_w= \sqrt{\frac{67500}{11}}$
$v_w=78$

We can conclude that the speed of the wind is 78 mi/hr.

6 0

3 years ago

a point on a perpendicular bisector is 7cm from each endpoint of the bisected segment and 5cm from the point of intersection. Wh

love history [14]

9514 1404 393

Answer:

about 9.80 cm

Step-by-step explanation:

The length of half the segment (h) can be found from the Pythagorean theorem:

h² +5² = 7²

h² = 7² -5² = 49 -25 = 24

h = √24 = 2√6