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Viefleur [7K]
2 years ago
13

-n+(-3)+3n+5 Combing like terms :

Mathematics
1 answer:
Karolina [17]2 years ago
3 0

Answer: 2n + 2 should be your answer

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Statistics urgent<br> Use n=6 and p=0.85 to complete
Bas_tet [7]

Answer:

I cant see anything on that picture. Wish I could help bruv. I wish I could help. It's a shame innit.

7 0
3 years ago
Solve the system of equations by substitution. What is the solution for x? 2x + y = 1 4x + 2y = −1
lapo4ka [179]
Solve your system of equations.

2x+y=1;4x+2y=−1

Solve 2x+y=1 for y:

2x+y+−2x=1+−2x(Add -2x to both sides)

y=−2x+1

Substitute (−2x+1) for y in 4x+2y=−1:

4x+2y=−1

4x+2(−2x+1)=−1

2=−1(Simplify both sides of the equation)

2+−2=−1+−2(Add -2 to both sides)

0=−3

Answer: No solution. C)

8 0
3 years ago
Read 2 more answers
a candy bar box is in the shape of a triangular prism. The volume of the box is 1200 cubic centimetres. The base is 10 centimetr
Elina [12.6K]

Answer:

Step-by-step explanation:

V=\frac{HBL}{2}\\ \\ H=\frac{2V}{BL}\\ \\ H=\frac{2(1200)}{10(20)}\\ \\ H=12cm

7 0
2 years ago
20 POINTS ANSWER FAST
Bond [772]

Answer

False

Step-by-step explanation:

8 0
3 years ago
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Cubed root x cubed root x2​
Yuki888 [10]

Answer:

Final answer is \sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x.

Step-by-step explanation:

Given problem is \sqrt[3]{x}\cdot\sqrt[3]{x^2}.

Now we need to simplify this problem.

\sqrt[3]{x}\cdot\sqrt[3]{x^2}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}

Apply formula

\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}

so we get:

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x

Hence final answer is \sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x.

7 0
3 years ago
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