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4 years ago

Please help! What is the value of X?

Mathematics

1 answer:

castortr0y [4]4 years ago

4 0

Angles are vertical opposite so they would be equal.
2x + 2 = 3x - 52
3x - 2x = 52 + 2
x = 54

In short, Your Answer would be 54

Hope this helps!

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Find all zeros.

Aleks [24]

Answer:

look at degree of polynomial

Step-by-step explanation:

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3 years ago

Anyone know this? that box is just when I get the answr

Ahat [919]

Answer:

1/10

Step-by-step explanation:

1/5 x 1/2

1/10

Bye , hope this helps!

Give brainliest !

5 0

3 years ago

Read 2 more answers

-(7-9)z-6z=8(-6+2) solve for z

Sergeeva-Olga [200]

Answer:

(-(7-9))(z)-6z= 8(-6+2)

Simplify

-(-2)(z) -6z =8(-6+2)

Multiply 8*-6 and 8*2

2z -6z= -48+16

Combine like terms

2z+-6z =-32

-4z=-32

Divide both sides by -4

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Step-by-step explanation:

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3 years ago

3/17=k/10 what is the missing value

Tcecarenko [31]

I think it is about 1.8 if you round it to the nearest tenth and cross multiply

3 0

3 years ago

2.75 Cards are dealt, one at a time, from a standard 52-card deck. a If the first 2 cards are both spades, what is the probabili

makvit [3.9K]

Answer:

The answer is "0.00842".

Step-by-step explanation:

Due to these cards, a conventional 52-card set is dealt one at a time.

While the first two cards are spades, then it would be expected that we could find the next three cards tents, too.

Let A draw 2 spade cards out of 52 cards. Let B become the occasion to draw 3 spade cards the remaining 50 end cards $(A \cap B)$ was its case where a 52 card spade is chosen $(2+3)=5$ .

The number of plots drawn first from 13 regular 52-cerd decks equals number of, $n(A)=\binom{13}{2}$ probability of event $A, PA =\frac{\binom{13}{2}}{\binom{52}{2}}$

Furthermore, the multitude of possibilities we can pull from of the remaining 11 spades of the previous 3 cards is 50-card decks standard, $n (B) = \binom{11}{3}$ then, the probability of event, $B, P(B)=\frac{\binom{11}{3}}{\binom{50}{3}}$

The chances of five cards being drawn (three spades and two spades),

$P(B \cap A)=\frac{\binom{13}{5}}{\binom{52}{5}}$

Then there is the chance that the next three cards will be picked if the first two are both pads, $P(\frac{B}{A})$

$\to P(\frac{B}{A}) \\\\ =\frac{P(B\cap A )}{P(A)}\\\\=\frac{\frac{\frac{13}{5}}{\frac{52}{2}}}{\frac{\frac{13}{2}}{\frac{52}{2}}}\\\\= 0.00842$

4 0

3 years ago