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S_A_V [24]
3 years ago
11

Solve the following system of equations using the

Mathematics
2 answers:
Alja [10]3 years ago
6 0

Answer:

The solutions are 3 and -5

Step-by-step explanation:

x+3=x^2+3x-12

x^2+2x-15

=(x-3)(x+5)

Alborosie3 years ago
3 0

y = x² + 3x - 12

y = x + 3

Substitute the second value of y into the first equation, and you'll have

x + 3 = x² + 3x - 12

Now solve

0 = x² + 2x - 15

0 = (x + 5)(x - 3)

x = - 5, x = 3

With -5, you'll get y = - 2

With 3, you'll get y = 6

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Find the number of positive integers less than 100,000 whose digits are among 1, 2, 3, and 4. Hint consider the
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Answer:

<em>1364</em> is the number of possibilities for positive integers less than 1,00,000.

Step-by-step explanation:

<em>1. 5 digit numbers:</em>

We have 5 places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 5 digit numbers:

4 \times 4 \times 4 \times 4 \times 4\\ \Rightarrow 1024

<em>2. 4 digit numbers:</em>

We have 4 places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 4 digit numbers:

4 \times 4 \times 4 \times 4 \\ \Rightarrow 256

<em>3. 3 digit numbers:</em>

We have 3 places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 3 digit numbers:

4 \times 4 \times 4 \\ \Rightarrow 64

<em>4. 2 digit numbers:</em>

We have <em>2</em> places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 2 digit numbers:

4 \times 4 \\ \Rightarrow 16

<em>5. 1 digit numbers:</em>

We have 1 place here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 1 digit numbers:

4

We can add all the above possibilities to find the total.

So,  number of positive integers less than 100,000 whose digits are among 1, 2, 3, and 4 = 1024 + 256 + 64 + 16 + 4 = <em>1364</em>

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3 years ago
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