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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.

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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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