All categories

3 years ago

An integer is 15 more than 2 tomes another. If the product of two integers is -27 then find the integers

Mathematics

1 answer:

andre [41]3 years ago

7 0

Answer:

x = 9 and y = -3

Step-by-step explanation:

Let's say the first integer is x and the second one is y. Then, if we put the words into mathematical expressions, we get:

x = 15 + 2y

xy = -27

We can use substitution to solve this system of equations. We put 15 + 2y in for x in xy = -27:

(15 + 2y)*y = -27 ⇒ 2y² + 15y = -27 ⇒ 2y² + 15y + 27 = 0

Now, we have a quadratic to solve, so after some trial and error, we get:

(2y + 9)*(y + 3) = 0

Setting each of these to 0, we get two possible answers for y:

y = -9/2 or y = -3.

However, it says that y is an integer, so it can't be -9/2 and must be -3. We use this to find x:

x = 15 + 2*(-3) = 15 - 6 = 9

So, the two numbers are x = 9 and y = -3.

You might be interested in

PLEASE HELP ASAP PLEASE WILL GIVE BEST RATE AND BRAINLEST PLEASE

natima [27]

Answer:

I think it's 145.225 sq in.

Step-by-step explanation:

2.5 x 2 x 3.14 = 15.7

15.7 x 9.25 = 145.225

8 0

3 years ago

Read 2 more answers

F(x)=x^3+9x^2+23x + 15

Airida [17]

This is what I got hope it helps!

4 0

3 years ago

27^4–9^5 3^9 is divisible by 25

hichkok12 [17]

Answer:

Step-by-step explanation:

If you mean

27^4 - 9^5·3^9 = -1161730026

it has no factors of 5, so cannot be divisible by 25.

If you mean ...

(27^4 -9^5)/3^9 = ((3^3)^4 -(3^2)^5)/3^9 = 3^10(3^2 -1)/3^9 = 3(9-1) = 24

it is not divisible by 25, either.

6 0

3 years ago

Make a ten or a hundred to add mentally 339+123

lianna [129]

9+3=12
30+20=50
300+100=400
Then add.
12+50+400=462
<span>So 339+123=462</span>

5 0

3 years ago

I GIVE BRAINLIEST PLS HELP

ioda

The answer is 8
Here's why:
${ ( \frac{( {6}^{7}) \times ( {3}^{3}) }{( {6}^{6}) \times ( {3}^{4} ) } )}^{3} = \\ ( \frac{6}{3} ) ^{3} = \\ \frac{216}{27} = 8$
The exponents are subtracted one from another when divided.
$\frac{ a ^{b} }{ {a}^{c} } = {a}^{b - c}$
We can look at the problem this way:
$( \frac{6^{7} }{6 ^{6} } \times \frac{3^{3} }{ {3}^{4} } ) = (6^{7 - 6} \times {3}^{3 - 4} ) = \\ ({6}^{1} \times {3}^{ - 1} )$
Since we have the power of -1 on the 3, we apply this rule:
${a}^{ - b} = \frac{1}{ {a}^{b} }$
Also this rule because we have the power of 1 on the 6:
${a}^{1} = a$
Then we get this:
$(6 \times \frac{1}{3} )^{3} = ( \frac{6}{3} )^{3}$
We apply the rule:
$( \frac{a}{b}) ^{c} = \frac{ {a}^{c} }{ {b}^{c} }$
We get this:
$\frac{{6}^{3} }{ {3}^{3} } = \frac{216}{27} = 8$

4 0

4 years ago