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dalvyx [7]
3 years ago
7

The product of a negative 11 cubed and negative 11 squared

Mathematics
1 answer:
bekas [8.4K]3 years ago
5 0

Answer:

121 and 1331

Step-by-step explanation:

Because 11x11=121 and 11x11x11=1331

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9th grade Quizziz help please
jenyasd209 [6]

Answer:

a²+7a+10

a²+5a+2a+10

a(a+5)+2(a+5)

(a+5)(a+2)

a+2 is required factor

7 0
3 years ago
Can you solve this :-)? Identify the equation that means "eleven less than x is twenty-three" (A) x = 23 - 11. (B) x + 11 = 23.
mart [117]
Hey user!!

your answer is here (-:

given :- eleven less than x ( x - 11 ) = twenty three

final answer = x - 11 = 23

correct option is D. x - 11 = 23

if u need the value of x too, then Answer is :-

x - 11 = 23

x = 23 + 11

x = 34

cheers!!
3 0
3 years ago
Solve for Y to get the equation into slope-intercept form and then state the slope and state the slope and state the y-intercept
valentinak56 [21]
-8-4y=-5x
-4y=-5x+8
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3 years ago
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james has a job with a current salary of $42,000. he wants to earn a collage degree that will take him 2 years to complete and i
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It will take him ten years.
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Anyone can help me solve this equation using cross multiplying
Natali5045456 [20]

9514 1404 393

Answer:

  x = 1 or 5

Step-by-step explanation:

The notion of "cross-multiplying" is the idea that the numerator on the left is multiplied by the denominator on the right, and the numerator on the right is multiplied by the denominator on the left. This looks like ...

  \displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\ \longrightarrow\ (x-1)(3x-1)=(7)(2x-2)

Then the solution proceeds by eliminating parentheses, and solving the resulting quadratic equation.

  3x^2-4x+1=14x-14\\\\3x^2-18x+15=0\qquad\text{subtract $14x-14$}\\\\x^2-6x+5=0 \qquad\text{divide by 3}\\\\(x-1)(x-5)=0\qquad\text{factor}\\\\x\in\{1,5\}

_____

<em>Comment on "cross multiply"</em>

Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes <em>what the result looks like</em>. It doesn't adequately describe how you get there. The <em>one and only rule</em> in solving Algebra problems is "<em>whatever is done to one side of the equation must also be done to the other side of the equation</em>." If you multiply one side by one thing and the other side by a different thing, you are violating this rule.

What looks like "cross multiply" is really "<em>multiply by the product of the denominators</em> and cancel like terms from numerator and denominator." Here's what that looks like with the intermediate steps added.

  \displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\\\\\frac{x-1}{7}\times7(3x-1)=\frac{2x-2}{3x-1}\times7(3x-1)\\\\(x-1)(3x-1)=(2x-2)(7)\qquad\textit{looks like}\text{ cross multiply}

8 0
3 years ago
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