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2 years ago

Does anyone have remind ill give u a code

Mathematics

2 answers:

Tanya [424]2 years ago

7 0

Answer:

can i has code

Step-by-step explanation:

i can give u memes

NeTakaya2 years ago

6 0

Answer:

Step-by-step explanation:

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In a 30° - 60° - 90° right triangle, the hypotenuse has a length of 8. what is the length of the longer leg?

Tatiana [17]

Sin 30° = 1/2 = x/8
then x = 4

sin 60° = √3/2 = y/8
then y = 4√3

so the length of the longer leg is 4√3

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3 years ago

PLZZ HELPP IM FAILLING THIS CLASS

Nataly_w [17]

Answer:

Well thats not good

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2 years ago

The table below is representing direct<br> variation.

valentina_108 [34]

Where is the table? I can’t help unless I can see the table

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2 years ago

Based on data from a statistical abstract, only about 16% of senior citizens (65 years old or older) get the flu each year. Howe

SSSSS [86.1K]

Answer: The required probability is 2.24%.

Step-by-step explanation:

Since we have given that

Percentage of people are senior citizens ( 65 years old or older ) = 14%

Percentage of people are under 65 years old = 100-14 = 86%

Probability that senior citizens get the flu each year = 16%

Probability that under 65 years old get the flu each year = 30%

So, Probability that a person selected at random from the general population is senior citizen who get the flu this season is given by

$P(S\cap F)={P(F|S)\times P(S)\\\\P(S\cap F)=0.16\times 0.14\\\\P(S\cap F)=0.0224\\\\P(S\cap F)=2.24\%$ Hence, the required probability is 2.24%.

3 0

2 years ago

Read 2 more answers

Perform the indicated operations. Write the answer in standard form, a+bi.<br> 5-3i / -2-9i

Vsevolod [243]

$\huge \boxed{\mathfrak{Answer} \downarrow}$

$\large \bf\frac { 5 - 3 i } { - 2 - 9 i } \\$

Multiply both numerator and denominator of $\sf \frac{5-3i}{-2-9i} \\$ by the complex conjugate of the denominator, -2+9i.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)}) \\$

Multiplication can be transformed into difference of squares using the rule: $\sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$ .

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}}) \\$

By definition, i² is -1. Calculate the denominator.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85}) \\$

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

$\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85}) \\$

Do the multiplications in $\sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right)$ .

$\large \bf \: Re(\frac{-10+45i+6i+27}{85}) \\$

Combine the real and imaginary parts in -10+45i+6i+27.

$\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85}) \\$

Do the additions in $\sf-10+27+\left(45+6\right)i$ .

$\large \bf Re(\frac{17+51i}{85}) \\$

Divide 17+51i by 85 to get $\sf\frac{1}{5}+\frac{3}{5}i \\$ .

$\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i) \\$

The real part of $\sf \frac{1}{5}+\frac{3}{5}i \\$ is $\sf \frac{1}{5} \\$ .

$\large \boxed{\bf\frac{1}{5} = 0.2} \\$

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2 years ago