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iris [78.8K]
3 years ago
6

I need help please help ​

Mathematics
1 answer:
RSB [31]3 years ago
4 0

Answer:

113+67=180

53+127=180

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Simplify the following expression 6X2 plus 8X -3 minus 3X2-6
Murrr4er [49]

Answer:

-24

Step-by-step explanation:

I´m guessing by the big x sign you want me to multiply so the first step in the equation is to multiply because of PEMDAS so(6*2)+(8*-3)-3*2-6,

so 6*2=12,8*-3=-24,-3*2=-6

12+-24-6-6,

12-24=-12,-6+-6=-12

-12+-12=-24

7 0
3 years ago
There are some beads in the box. 40% of them were blue and the rest were red. Adam used all the blue beads and 25% of the red be
Shalnov [3]

Answer:

65% of the beads were used to make the bracelet.

Step-by-step explanation:

40% of the beads were blue which means 60% of the beads were red. If all the blue beads were used that means 40% was used so you add the 25% of the red beads that were used which is why the answer is 65%.

7 0
3 years ago
Paula is a banker who earns $15 per hour. If Paula worked 63 hours at her job, how much money did she earned? Please answer this
Mariana [72]
Paula made $945 if you multiply $15 x 63
8 0
3 years ago
Read 2 more answers
Z is..? The problem is attached as a png!
VikaD [51]

Answer:

Step-by-step explanation:

What i would do is multiply both sides by 2z + 1

so you get z-6=20z+10

-19z=16

z=-16/19

5 0
3 years ago
Read 2 more answers
If f(x)=2−x12 and g(x)=x2−9, what is the domain of g(x)÷f(x)?
Keith_Richards [23]
\bf \begin{cases}
f(x)=2-x^{12}\\
g(x)=x^2-9\\
g(x)\div f(x)=\frac{g(x)}{f(x)}
\end{cases}\implies \cfrac{x^2-9}{2-x^{12}}

now, for a rational expression, the domain, or "values that x can safely take", applies to the denominator NOT becoming 0, because if the denominator is 0, then the rational turns to undefined.

now, what value of "x" makes this denominator turn to 0, let's check by setting it to 0 then.

\bf 2-x^{12}=0\implies 2=x^{12}\implies \pm\sqrt[12]{2}=x\\\\
-------------------------------\\\\
\cfrac{x^2-9}{2-x^{12}}\qquad \boxed{x=\pm \sqrt[12]{2}}\qquad \cfrac{x^2-9}{2-(\pm\sqrt[12]{2})^{12}}\implies \cfrac{x^2-9}{2-\boxed{2}}\implies \stackrel{und efined}{\cfrac{x^2-9}{0}}

so, the domain is all real numbers EXCEPT that one.
4 0
3 years ago
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