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Lubov Fominskaja [6]
3 years ago
10

Please help me please please. I am stuck very badly please. I will do anything that you want but please help me please please.

Mathematics
1 answer:
Lyrx [107]3 years ago
5 0

Answer:

varies

Step-by-step explanation:

if you put a x or dot in the middle it will count as a dot/x in both circles so put the left over number or numbers there

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I might be wrong but x should be -360

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PLEASE ANSWER MY QUESTION I AM OFFERING 60 POINTS AND I REALLY NEED THE ANSWER
Irina-Kira [14]
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Ratio 11 to 6 in perimeter
Darina [25.2K]

Answer: the answer will be 6/4

Step-by-step explanation:

6 0
3 years ago
Hi need help on hw plz help me out my questions is - a rectangle has a length of 14cm and width 9.5cm the length increased by 6%
sweet-ann [11.9K]

Answer:

Part a) 21.9\%

Part b) 82.03\%

Step-by-step explanation:

we know that

The original area of rectangle is equal to

A=(14)(9.5)\ cm^{2}

so

if  the length is increased by 6% and the width is by 15%

then

the new area is equal to

A=[1.06(14)][1.15(9.5)]\ cm^{2}

A=(1.06*1.15)(14)(9.5)\ cm^{2}

A=(1.219)(14)(9.5)\ cm^{2}

The new area is (1.219) times the original area

Part a) Calculate the percentage change in area of the rectangle

1.219-1=0.219

convert to percentage

0.219*100=21.9\%

Part b) what is the original area as a percentage of the increased area of the rectangle?

Divide the original area by the increased area

\frac{(14)(9.5)}{(1.219)(14)(9.5)} =\frac{1}{1.219} \\ \\=0.8203

Convert to percentage

0.8203*100=82.03\%

8 0
4 years ago
Solve the system of equations y = 40x2 and y = 19x + 3 algebraically.
ikadub [295]

We have been given a system of nonlinear equations.

\\y=40x^{2},y=19x+3

In order to solve this system we can first equate the two equations in order to get a quadratic in x.

\\40x^{2}=19x+3

Our next step is to bring all the terms on one side.

\\40x^{2}-19x-3=0

Now we have to solve this equation. We can solve it by factoring using the splitting middle term method.

\\40x^{2}-24x+5x-3=0

\\8x(5x-3)+(5x-3)=0

\\(8x+1)(5x-3)=0

Upon setting each of these factors equal to zero using zero product property we get

\\8x+1=0 \text{ or } 5x-3 = 0

Upon solving both these equations we get

\\x=-\frac{1}{8} \text{ or } x=\frac{3}{5}

Now we can substitute these values of x in the equation

y=19x+3

We get

\text{For }x=- \frac{1}{8} \Rightarrow y=19(- \frac{1}{8})+3= \frac{5}{8}\\

\text{For }x= \frac{3}{5}\Rightarrow y=19( \frac{3}{5})+3= \frac{72}{5}\\

Therefore, our final set of solutions are

(x,y)=(- \frac{1}{8},\frac{5}{8}) \text{ and } ( \frac{3}{5}, \frac{72}{5})


6 0
3 years ago
Read 2 more answers
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