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2 years ago

Help me please this is a multiple question

Mathematics

2 answers:

konstantin123 [22]2 years ago

7 0

The is a right triangle

Dmitrij [34]2 years ago

7 0

Answer:

This is a right triangle you can tell because a right triangle looks like a "L" on any side or what so ever

Step-by-step explanation:

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What are the solutions to the equation?

olga2289 [7]

Answer:

x = −10 and x = 4

Step-by-step explanation:

x2 + 6x = 40

Subtract 40 from each side

x^2 + 6x -40 =0

Factor, what 2 numbers multiply to -40 and add to 6

10 * -4 = -40 10+-4 = 6

(x+10) (x-4) = 0

Using the zero product property

x+10 =0 x-4=0

x=-10 x=4

6 0

3 years ago

Read 2 more answers

Prove the following identity.<br> [sin(x+y)+sin(x-y)]/[cos(x+y)+cos(x-y)] = tan(x)

tia_tia [17]

$\frac{sin(x + y) + sin(x - y)}{cos(x + y) + cos(x - y)} = tan(x)$
$\frac{(sin(x)cos(y) + cos(x)sin(y)) + (sin(x)cos(y) - cos(x)sin(y))}{(cos(x)cos(y) - sin(x)sin(y)) + (cos(x)cos(y) + sin(x)sin(y))} = tan(x)$
$\frac{(sin(x)cos(y) + sin(x)cos(y)) + (cos(x)sin(y) - cos(x)sin(y))}{(cos(x)cos(y) + cos(x)cos(y)) + (-sin(x)sin(y) + (sin(x)sin(y))} = tan(x)$
$\frac{2sin(x)cos(y)}{2cos(x)cos(y)} = tan(x)$
$\frac{sin(x)}{cos(x)} = tan(x)$

4 0

2 years ago

The probability distribution of the number of students absent on Mondays, is as follows: X 0 1 2 3 4 5 6 7 f(x) 0.02 0.03 0.26 0

alexgriva [62]

a) Add up all the probabilities $f(x)$ where $x>3$ :

$f(4)+f(5)+f(6)+f(7)=0.35$

b) The expected value is

$E[X]=\displaystyle\sum_xx\,f(x)=3.16$

Since X is the number of absent students on Monday, the expectation E[X] is the number of students you can expect to be absent on average on any given Monday. According to the distribution, you can expect around 3 students to be consistently absent.

c) The variance is

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

where

$E[X^2]=\displaystyle\sum_xx^2\,f(x)=11.58$

So the variance is

$V[X]=11.58-3.16^2\approx1.59$

The standard deviation is the square root of the variance:

$\sqrt{V[X]}\approx1.26$

d) Since $Y=7X+3$ is a linear combination of $X$ , computing the expectation and variance of Y is easy:

$E[Y]=E[7X+3]=7E[X]+3=25.12$

$V[Y]=V[7X+3]=7^2V[X]\approx78.13$

e) The covariance of X and Y is

$\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$

We have

$XY=X(7X+3)=7X^2+3X$

$E[XY]=E[7X^2+3X]=7E[X^2]+3E[X]=90.54$

Then the covariance is

$\mathrm{Cov}[X,Y]=90.54-3.16\cdot25.12\approx11.16$

f) Dividing the covariance by the variance of X gives

$\dfrac{\mathrm{Cov}[X,Y]}{V[X]}\approx\dfrac{11.16}{1.59}\approx0.9638$

5 0

3 years ago

Select the correct answer.

Molodets [167]

To answer this question you must know that there is a theorem called the "Two secants angle theorem", which establishes that the measure of an angle formed when two secants intersecting outside the circle is equal to 1/2 the difference of the two intercepted arcs.

Therefore, as you can see, the correct answer is the option B): B)<span>1/2 the difference of the intercepted arcs</span>

4 0

2 years ago

Mr jumbo borrowed 4000 Naira from a bank he agrees to pay the money back after two years with an interest of 9% per annum how mu

Zarrin [17]

Answer:

180.oo

Step-by-step explanation:

4000x2x9

20x9

180

8 0

2 years ago