All categories

3 years ago

Please help math experts need your help 15 points !

Mathematics

1 answer:

Alinara [238K]3 years ago

7 0

Answer:

12 = -3 - 3x

Step-by-step explanation:

Same concept as the last question I answered. First, let's derive a formula from the given model. So we have 3 negative x and 3 negative 1 which equal to 12 positive ones, this can be written as:

-3x - 3 = 12

which can be rewritten as

12 = -3 - 3x

You might be interested in

There are 12 contestants in a coloring contest. Three ribbons will be awarded. How many different ways could the ribbons be awar

Elanso [62]

Answer:

1320 ways

Step-by-step explanation:

Here we have a situation where 3 ribbons will be awarded to 3 of the 12 contestants. We can use the permutation formula, because the order of awarding the ribbons matters.

The permutation formula is:

$_{n}P_{k}=\dfrac{n!}{(n-k)!}$

n = 12

k = 3

Now that we have our variables, let's plug them into the formula.

$_{12}P_{3}=\dfrac{12!}{(12-3)!}$

$_{12}P_{3}=\dfrac{12!}{9!}$

$_{12}P_{3}=1320$

So there are 1320 different ways that the contestants will be awarded.

5 0

3 years ago

if you were to reflect figure ABCD over the line of reflection y=x, what would be the coordinate if A’ be?

Phantasy [73]

Answer:

The x coordinates and y coordinates will switch places with each other.

Step-by-step explanation:

A' coordinates will go from (x,y) to (y,x) so the coordinates will switch.

5 0

3 years ago

X² - kx+4-0<br>If the equation has<br>equal roots, then<br>b²-4ac=0<br>where a=1, b = -k and c=4

olga55 [171]

Answer:

k = ± 4

Step-by-step explanation:

Given

x² - kx + 4 = 0 ← in standard form

with a = 1, b = - k, c = 4

The equation has equal roots, thus the discriminant

b² - 4ac = 0, that is

(- k)² - (4 × 1 × 4) = 0

k² - 16 = 0 ( add 16 to both sides )

k² = 16 ( take the square root of both sides )

k = ± $\sqrt{16}$ = ± 4

3 0

3 years ago

For some civil cases, at least 9 of

Inessa [10]

Using the combination formula:

n! / (r!(n-r)!)

Where n is the number of people = 12 and r is the number of jurors = 9.

12! / (9!(12-9)!) = 220 combinations.

5 0

3 years ago

In how many ways you can form a combination of 3 numbers and 3 letters in a plate number if repetition is allowed

Lapatulllka [165]

Your plate numbers will be like

$xxxyyy$

where each $x$ is a number, and each $y$ is a letter. Assuming you can use all numbers and letters, you have 10 possible choices for every $x$ place (all the digits from 0 to 10) and 26 possible choices for every $y$ place (all the letters from a to z).

So, if you multiply all the possible choices, you have

$xxxyyy \to 10\cdot 10 \cdot 10 \cdot 26 \cdot 26 \cdot 26 = 10^3 \cdot 26^3 = (10\cdot 26)^3 = 260^3$

So, there is a total of

$260^3 = 17,576,000$

possible plate numbers with 3 letters and 3 numbers, if repetitions are allowed.

5 0

3 years ago