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2 years ago

13

Four students are competing in a spelling bee. The top three students will receive a medal. How many different ways can they fin

ish 1st, 2nd, and 3rd?

2 answers:

Gre4nikov [31]2 years ago

7 0

Answer:

<em>How many different ways can the letters of the word MATH be rearranged to form a four- ... 4 · 3 · 2 · 1 = 24 ways to spell a code worth with the letters MATH. ... example, coming in first means that you get the gold medal instead of the ... A group of four students is to be chosen from a 35-member class to represent the class.</em>

Step-by-step explanation:

GarryVolchara [31]2 years ago

5 0

Answer:

24 ways

Step-by-step explanation:

I just made the dude's answer above simpler.

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Ken grew 4/5 of an inch last year. sang grew3/8 of an inch . who grew more and by how much?

Licemer1 [7]

Sang grew more. and by 17/40 ( I could be wrong)

4 0

3 years ago

The coefficient of xy in the product of (4x2 + 2y) and (3x + y2) is

dedylja [7]

(4x2 + 2y)(3x+y^2)

= 12x^3 + 4x^2y^2 + 6xy + 2y^3

coefficient of xy is ----> 6

8 0

3 years ago

Read 2 more answers

Suppose the graph of a cubic polynomial function has the same zeroes and passes through the coordinate (0, -5). Write the equati

amid [387]

In this we know all three zeros and one point from which the graph pass.
So we will let specific cubic polynomial function of the form
$f(x) = a(x - x_1)(x-x_2)(x-x_3)$

As we know zeros are that point where we will get value of function equal to zero. So it is basically in form $(x_n , 0)$

SO in given question zeros are (2 , 0) , (3, 0) and (5,0)
So we can say $x_1 = 2 , x_2 = 3 , x_3 = 5$

So required equation is
$f(x) = a (x-2)(x-3)(x-5)$
              $= a[(x^2 - 2x - 3x + 6)(x-5)]$
              $= a[(x^2 - 5x+6)(x-5)]$
              $= a(x^3 - 5x^2 + 6x- 5x^2 + 25x - 30)$
              $= a(x^3 - 10x^2+31x-30)$
Now we have one point (0 , -5) from which graph passes.
So we say at x = 0 , f(x) = -5
$-5= a (0-0+0 - 30)$
$-5 = -30a$
$a = \frac{-5}{-30} = \frac{1}{6}$
So required equation of cubic polynomial is
$f(x) = \frac{1}{6}(x^3 -10x^2+31x-30)$

For finding y - intercept we simply plugin x = 0 in given equation.
As we know at x = 0 , value of function is -5.
So y - intercept is -5.

4 0

2 years ago

Read 2 more answers

Subtract (7.8 - 5.1t) from (2.8 - 3.2t). Use

tankabanditka [31]

Answer: we can use the commutative property to show the difference in the following ways:-

a. (-3.2t + 2.8) - (-5.1t + 7.8)

b. 1.9t - 5

c. (-3.2t + 2.8) - (-5.1t + 7.8)

Step-by-step explanation:

Commutative simply has to do with "moving around". In the case of addition, the commutative property rule is "x + y = y + x". When it comes to numbers, it my 7 + 8 = 8 + 7. In the case of multiplication, the commutative property rule is "m × n = n × m". So whenever commutative property is referred to, what is expected is "to move around". Commutative property rule can be executed as you pair, add or subtract like terms e.g 2a - 4 + 9a can be rewritten as 2a + 9a - 4 or 11a - 4

In the question, we were asked to subtract (7.8 - 5.1t) from (2.8 - 3.2t) and use the commutative property to show the difference in another way.

Subtracting (7.8 - 5.1t) from (2.8 - 3.2t) will give:-

(2.8 - 3.2t) - (7.8 - 5.1t) (This is the original expression)

Removing the bracket:

2.8 - 3.2t - 7.8 + 5.1t

= 5.1t - 3.2t + 2.8 - 7.8 (we have now shown the difference in another way)

Subtracting the like terms, will give: 1.9t - 5 (This is also the difference shown in another way).

Recall that the initial expression was

(2.8 - 3.2t) - (7.8 - 5.1t).

We can also use the commutative property rule to show the difference in the following way:-

(-3.2t + 2.8) - (-5.1t + 7.8)

By so doing, the figures have been rearranged without altering or affecting the outcome of the expression.

4 0

2 years ago

Which values of x and y would make the following expression represent a real number? (4+5i)(x+yi)

algol [13]

Answer:

So x=4 and y=-5 would work.

Step-by-step explanation:

When you multiply complex conjugates, you get a real result.

Example:

$(a+bi)(a-bi)=a^2-b^2i^2=a^2+b^2$

I replace $i^2$ with -1 since $i^2=-1$ .

$a^2+b^2$ is a real number (there is no imaginary part, no i).

So what is the conjugate of 4+5i?

4-5i

So x=4 and y=-5 would work.

A multiple of the complex conjugate would work as well.

$(a+bi)(ca-cbi)$

$c(a+bi)(a-bi)$

$c(a^2-b^2i^2)$

$c(a^2+b^2)$

This is still a real number; there is no imaginary part,no i.

5 0

2 years ago