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3 years ago

Partial products for 315 4

Mathematics

1 answer:

max2010maxim [7]3 years ago

6 0

Do 315 and start to multiply from 4 and you will find your answers

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Solve the system:<br> 4x – 3y = -21<br> X=-y-14

Liula [17]

Answer:

The solution for the system of equations is

x=2,y=−2

Explanation:

4x−3y=14.....equation (1)

y=−3x+4.....equation (2)

Solving by substitution

Substituting equation 2 in equation 1

4x−3⋅(−3x+4)=14

4x+9x−12=14

13x=14+12

13x=26

x=2

Finding y by substituting x in equation 1

4x−3y=14

4⋅2−3y=14

8−3y=14

8−14=3y

3y=−6

y=−2

4 0

3 years ago

Read 2 more answers

A French class has a total of 31 students. The number of males is 7 more than the number of females. How many males and how many

Inessa05 [86]

Answer:

Step-by-step explanation:

5 0

3 years ago

Use exterior angles to show that the angle sum of quadrilateral PQRS is 360°.

Shkiper50 [21]

Answer:

I don't really know but one triangle is 180 degrees so two added together is 360 degrees...

but ohh I remembered

a=c+d (since a+b=180 and c+d+b=180)

e=g+h (since e+f=180 and g+h+f= 180)

and the fact is that c+d+g+h+b+f=the angle sum of quadrilateral PQRS

a+e+b+f= the angle sum of quadrilateral PQRS (by substitutions)

a+e+b+f=360 (four angles are from one origin so it's 360 degrees )

8 0

3 years ago

Suppose you pick two cards from a deck of 52 playing cards. What is the probability that they are both queens?

photoshop1234 [79]

Answer:

0.45% probability that they are both queens.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

The combinations formula is important in this problem:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Desired outcomes

You want 2 queens. Four cards are queens. I am going to call then A,B,C,D. A and B is the same outcome as B and A. That is, the order is not important, so this is why we use the combinations formula.

The number of desired outcomes is a combinations of 2 cards from a set of 4(queens). So

$D = C_{4,2} = \frac{4!}{2!(4-2)!} = 6$

Total outcomes

Combinations of 2 from a set of 52(number of playing cards). So

$T = C_{52,2} = \frac{52!}{2!(52-2)!} = 1326$

What is the probability that they are both queens?

$P = \frac{D}{T} = \frac{6}{1326} = 0.0045$

0.45% probability that they are both queens.

4 0

3 years ago

Need answer to this math problem

Elis [28]

The slope of this line is the difference of the y values divided by the difference of the x values. Solving this give 4/-3. Then, you already know the b in y = mx+b, which is 2, because you already have the value (0, 2). So, the answer is y = $-\frac{4}{3}$ x + 2

8 0

3 years ago

Read 2 more answers