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3 years ago

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

Mathematics

2 answers:

____ [38]3 years ago

6 0

Answer:

The solutions are $-2+ 2\sqrt{6}\texttt{ and}-2- 2\sqrt{6}$

Step-by-step explanation:

We have to find solution of x²+4x+10.

Solution is given by

$\frac{-4\pm \sqrt{4^2-4\times 1\times 10}}{2\times 1}=\frac{-4\pm \sqrt{16-40}}{2}=-2\pm 2\sqrt{6}$

So the solutions are $-2+ 2\sqrt{6}\texttt{ and}-2- 2\sqrt{6}$

alexira [117]3 years ago

4 0

The solution is 4x+2x=6x then 6x+10=solution

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Select the best answer. tan x tan x/2=

Colt1911 [192]

Tan 2 x-2 tan x=0
Tan x (tan x-2 ) =0
A tan x=0 —> x=0 and x =K π
B tan x=2 —> x=63.43 +k180 deg

4 0

3 years ago

A triangle with a height of 5cm and a base length of 12cm what is the perimeter

Lisa [10]

Answer: 5+5+12+12

Step-by-step explanation:

7 0

3 years ago

The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, So

tatyana61 [14]

Answer:

The probability that at least 13 flights arrive late is 2.5196 $\times 10^{-6}$ .

Step-by-step explanation:

We are given that Southwest Air had the best rate with 80 % of its flights arriving on time.

A test is conducted by randomly selecting 18 Southwest flights and observing whether they arrive on time.

The above situation can be represented through binomial distribution;

$P(X = x) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.........$

where, n = number of trials (samples) taken = 18 Southwest flights

r = number of success = at least 13 flights arrive late

p = probability of success which in our question is probability that

flights arrive late, i.e. p = 1 - 0.80 = 20%

Let X = Number of flights that arrive late.

So, X ~ Binom(n = 18, p = 0.20)

Now, the probability that at least 13 flights arrive late is given by = P(X $\geq$ 13)

P(X $\geq$ 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)

= $\binom{18}{13}\times 0.20^{13} \times (1-0.20)^{18-13}+ \binom{18}{14}\times 0.20^{14} \times (1-0.20)^{18-14}+ \binom{18}{15}\times 0.20^{15} \times (1-0.20)^{18-15}+ \binom{18}{16}\times 0.20^{16} \times (1-0.20)^{18-16}+ \binom{18}{17}\times 0.20^{17} \times (1-0.20)^{18-17}+ \binom{18}{18}\times 0.20^{18} \times (1-0.20)^{18-18}$

= $\binom{18}{13}\times 0.20^{13} \times 0.80^{5}+ \binom{18}{14}\times 0.20^{14} \times 0.80^{4}+ \binom{18}{15}\times 0.20^{15} \times 0.80^{3}+ \binom{18}{16}\times 0.20^{16} \times 0.80^{2}+ \binom{18}{17}\times 0.20^{17} \times 0.80^{1}+ \binom{18}{18}\times 0.20^{18} \times 0.80^{0}$

= 2.5196 $\times 10^{-6}$ .

7 0

3 years ago

Sec squared 55 - tan squared 55

sergejj [24]

Answer:

sec squared 55 – tan squared 55 = 1

Explanation:

Given, sec square 55 – tan squared 55

We know that,

$\sec \Theta=\frac{\text {hypotenuse}}{\text {base}}$

And,

$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$

where Ө is the angle

Substituting the values

$\left(\frac{\text {hypotenuse}}{\text {base}}\right)^{2}-\left(\frac{\text { perpendicular }}{\text {base}}\right)^{2}$

Solving,

$\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}$

According to Pythagoras theorem,

$\text { (hypotenuse) }^{2}-\text { (perpendicular) }^{2}=(\text { base })^{2}$

Putting this in the equation;

squared 55 - tan squared 55 =

$\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}=\frac{(\text {base})^{2}}{(\text {base}) *(\text {base})}=1$

Therefore, sec squared 55 – tan squared 55 = 1

6 0

3 years ago

A pipe is leaking at 1.5 cups per day. about how many gallons per week is the pipe leaking

Alenkinab [10]

Remember that there are 16 cups in a gallon.

The pipe leaks 1.5 cups each day, so in one week, or 7 days, the cup leaks 1.5 times 7 cups, which equates to 10.5 cups per week.
To find this number in gallons, divide 10.5 by 16.
This equals .65625 gallons per week.
Hope this helps!
~Chrys

8 0

3 years ago