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Ray Of Light [21]
3 years ago
12

Which of the following is the solution set of the equation shown below? 2x(x+3)-5(x+3)=0​

Mathematics
2 answers:
Eddi Din [679]3 years ago
8 0

Answer:

(2)

Step-by-step explanation:

Kruka [31]3 years ago
7 0

Answer:

Step-by-step explanation:

3

x

+

7

=

0

Subtract  

7

from both sides of the equation.

3

x

=

−

7

Divide each term by  

3

and simplify.

Tap for more steps...

x

= 2 1/3

−

7

3

The result can be shown in multiple forms.

Exact Form:

x

=

−

7

3

Decimal Form:

x

=

−

2.

¯

3

Mixed Number Form:

x

=

−

2

1

3

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Ierofanga [76]

5 is the answer to your question.  

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Find y when x=9 if y varies directly as x, and y=245 when x=7
Aleonysh [2.5K]
Since y varies directly as x, you know that y=245 and x=7 can be used as a proportion, 245/7.  When you multiply this proportion to 9, you get 315.  This is the value of y when x=9.  

y=315
8 0
3 years ago
FIRST RIGHT GETS BRAINLIEST
ki77a [65]

Answer:

Area = 90 units^2

Step-by-step explanation:

For the area of a triangle, we can simply use the following equation:

Area = ( 1 / 2 ) * Base * Height

We are given a base of 12 and a height of 15, so we can simply plug those values into the formula to find the area.

Area = ( 1 / 2 ) * Base * Height

Area = ( 1 / 2 ) * 12 * 15

Area = 6 * 15

Area = 90

Hence, the area of our triangle is 90 units^2.

Cheers.

6 0
4 years ago
Find the values of A and B<br> PLEASE HELP GIVING BRAINLIEST
erastovalidia [21]

Factoring the roots, the values of A and B are given as follows:

  • A = 3z.
  • B = 2.

<h3>What are the values of A and B?</h3>

The expression is:

\sqrt{\frac{45\alpha z^3}{40y}} = \frac{A\sqrt{\alpha z}}{B\sqrt{2y}}

We have that:

\frac{45}{40} = \frac{9}{8}

Hence:

\sqrt{\frac{45\alpha z^3}{40y}} = \sqrt{\frac{9\alpha z^3}{8}} = \sqrt{\frac{9\alpha z^2 \times z}{4 \times 2}} = \frac{A\sqrt{\alpha z}}{B\sqrt{2y}}

Hence, comparing the last two equalities:

  • A = 3z.
  • B = 2.

More can be learned about the factoring of roots at brainly.com/question/24380382

#SPJ1

7 0
2 years ago
Square root of 144 X^7 Y^5
LekaFEV [45]
The first step to solving this is to factor out the first perfect square
\sqrt{12^{2} x^{7} y^{5}   }
now factor out the second perfect square
\sqrt{ 12^{2} x^{6} X x  y^{5}  }
then factor out the second perfect square 
\sqrt{ 12^{2}  x^{6} X x  y^{4} X y}
the root of a product is equal to the product of the roots of each factor
\sqrt{ 12^{2} } \sqrt{ x^{6} } \sqrt{ y^{4} } \sqrt{xy}
reduce the index of the radical and exponent with 2 of the first square root
12\sqrt{ x^{6} } \sqrt{ y^{4} } \sqrt{xy}
reduce the index of the radical and exponent with 2 of the second square root
12x³\sqrt{ y^{4} } \sqrt{xy}
reduce the index of the radical and exponent with 2 of the third square root
12x³y²\sqrt{xy}
this means that the correct answer to your question is 12x³y²\sqrt{xy} .
let me know if you have any further questions
:)
4 0
3 years ago
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