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kodGreya [7K]
4 years ago
6

Sharon uses 72 centimeters of ribbon to wrap gifts she uses 24 centimeters of her total ribbon to wrap a big gift she uses the r

emaining ribbon for 6 small gifts how much ribbon will she use for each gift if she uses the same amount on each
Mathematics
2 answers:
xxMikexx [17]4 years ago
4 0

Answer:

8

Step-by-step explanation:

tekilochka [14]4 years ago
3 0

Answer:

Step-by-step explanation:

Length of the remaining ribbon = 72 - 24 = 38cm

The length of ribbon used for each gift = 48 / 6

             = 8cm

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7,000 is 10 times as much as 
Llana [10]
I think the answer is 700 as 700*10=7,000.
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4 years ago
What is the missing step in solving the inequality 5 – 8x < 2x + 3? Add 2x to both sides of the inequality. Subtract 8x from
kvasek [131]

Answer:

⇒ Add 8x to both sides of the inequality

⇒ x>1/5

Step-by-step explanation:

First, you subtract by 5 from both sides of equation.

5-8x-5<2x+3-5

Solve.

-8x<2x-2

Then subtract by 2x from both sides of equation.

-8x-2x<2x-2-2x

Solve.

-10x<-2

Multiply by -1 from both sides of equation.

(-10x)(-1)>(-2)(-1)

Solve.

10x>2

Divide by 10 from both sides of equation.

10x/10>2/10

Solve to find the answer.

2/10=10/2=5 2/2=1=1/5

x>1/5 is final answer.

Hope this helps!

5 0
4 years ago
Read 2 more answers
Witch of the following best describes the slope of the line below ​
ankoles [38]
Answer- undefined slope
7 0
3 years ago
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Find the difference using suitable identity (10.3 × 9.7) – (10.7 × 9.3)
finlep [7]

Answer:

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5 0
3 years ago
I need help solving this problem. <img src="https://tex.z-dn.net/?f=4x%20cos%20%5E%7B-1%7D%20%282x%2B4%29-%20%5Csqrt%7B3-3%20x%5
Gnesinka [82]
Given:
f(x)=4xcos^{-1}(2x+4)-\sqrt{3-3x^2}

Using
\frac{d}{dx}cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}
we derive
\frac{d}{dx}4xcos^{-1}(2x+4)
=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}

Similarly, using
\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}
we derive
\frac{d}{dx}(-\sqrt{3-3x^2})
=\frac{3x}{\sqrt{3-3x^2}}

Therefore, the derivative is
f'(x)=\frac{d}{dx}(4xcos^{-1}(2x+4)-\sqrt{3-3x^2})
=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}+\frac{3x}{\sqrt{3-3x^2}}
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3 years ago
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