In the rectangle below, EI = 2x+6, FI= 5x-6, and m ZIFE = 490.

Mathematics

1 answer:

worty [1.4K]3 years ago

3 0

Answer:

gyufigaruofhasidjfdskcm

Step-by-step explanation:

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Please include equation and solution thanks

ehidna [41]

John is 31 and Jenny is 26

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2 years ago

Write an equation for the orbit of Saturn in the form x^2/a^2 + y^2/b^2 = 1

frozen [14]

Answer:

x²/2166784 +y²/2159989 = 1

Step-by-step explanation:

The relationship between the semi-axes and the eccentricity of an ellipse is ...

e = √(1 -b²/a²)

In order to write the desired equation we need to find 'b' from 'e' and 'a'.

<h3>semi-minor axis</h3>

Squaring the equation for eccentricity gives ...

e² = 1 -b²/a²

Solving for b², we have ...

b²/a² = 1 -e²

b² = a²(1 -e²)

<h3>ellipse equation</h3>

Using the given values, we find ...

b² = 1472²(1 -0.056²) = 2166784(0.996864) ≈ 2159989

The desired equation is ...

x²/2166784 +y²/2159989 = 1

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2 years ago

How do I solve this ?

noname [10]

The answer is around 29

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3 years ago

Your English teacher has decided to randomly assign poems for the class to read. The syllabus includes four poems by Shakespeare

Tresset [83]

The probability of event A and B to both occur is denoted as P(A ∩ B) = P(A) P(B|A). It is the probability that Event A occurs times the probability that Event B occurs, given that Event A has occurred.

So, to find the probability that you will be assigned a poem by Shakespeare and by Tennyson, let Event A = the event that a Shakespeare poem will be assigned to you; and let Event B = the event that the second poem that will be assigned to you will be by Tennyson.

At first, there are a total of 13 poems that would be randomly assigned in your class. There are 4 poems by Shakespeare, thus P(A) is 4/13.
After the first selection, there would be 13 poems left. Therefore, P(B|A) = 2/12
Based on the rule of multiplication,
P(A ∩ B) = P(A) P(B|A)P(A ∩ B) = 4/13 * 2/12
P(A ∩ B) = 8/156
P(A ∩ B) = 2/39

The probability that you will be assigned a poem by Shakespeare, then a poem by Tennyson is 2/39 or 5.13%.

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3 years ago

In a chess club the probability that Shaun will beat Mike is 3/8 .

romanna [79]

Answer:

$\frac{10}{56} = 0.1786$ probability that Shaun loses both games