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nlexa [21]
2 years ago
14

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Mathematics
1 answer:
ANTONII [103]2 years ago
6 0

Answer:

the answer is (3,4)

Step-by-step explanation:

because if you check wich quadrant it is in and you check the number you can see which one it is. x,y that is how it goes so 3,4 should be correct if you line them up.

You might be interested in
Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)
Paladinen [302]

Answer:

\frac{140}{3}, \frac{140}{3}, \frac{140}{3}

Step-by-step explanation:

Let the numbers be x, y\ and\ z.

Such that:

x + y + z = 140

Make z the subject

z = 140 -x - y

For their product to be maximum, we have:

f(x,y,z) = xyz

Substitute z = 140 -x - y in f(x,y,z) = xyz

f(x,y) = xy(140 - x - y)

Open bracket

f(x,y) = 140xy - x^2y - xy^2

Differentiate w.r.t x and y

f_x=140y - 2xy - y^2

f_y=140x - x^2 - 2xy

Since the products are maximum, then f_x = f_y = 0

For f_x=140y - 2xy - y^2

140y - 2xy - y^2 = 0

Factorize:

y(140 - 2x - y) = 0

Split

y = 0\ or\ 140 - 2x - y = 0

Make y the subject

y = 0\ or\ y = 140 - 2x

For f_y=140x - x^2 - 2xy

140x - x^2 - 2xy = 0

---------------------------------------------------

Substitute y = 0

140x - x^2 -2x*0 = 0

140x - x^2 = 0

Factorize

x(140 - x)= 0

x = 0\ or\ 140-x = 0

x = 0\ or\ x = 140

---------------------------------------------------

Substitute y = 140 - 2x

140x - x^2 - 2xy = 0

140x - x^2 - 2x(140 - 2x) = 0

140x - x^2 - 280x + 4x^2 = 0

Re-arrange

4x^2 -x^2 +140x - 280x = 0

3x^2 -140x = 0

Factor x out

x(3x - 140) = 0

Divide through by x

3x - 140 = 0

3x = 140

x = \frac{140}{3}

Recall that: y = 140 - 2x

y = 140 - 2 * \frac{140}{3}

y = 140 - \frac{280}{3}

Take LCM

y = \frac{140*3-280}{3}

y = \frac{140}{3}

Recall that:

z = 140 -x - y

z = 140 - \frac{140}{3} - \frac{140}{3}

Take LCM

z =  \frac{3 * 140- 140 - 140}{3}

z =  \frac{140}{3}

Hence, the numbers are:

\frac{140}{3}, \frac{140}{3}, \frac{140}{3}

8 0
3 years ago
Polygon F has an area of 36 square units. Aimar drew a scaled version of Polygon F and labeled it Polygon G. Polygon G has an ar
Free_Kalibri [48]

Answer:

1/3

Step-by-step explanation:

The area of Polygon GGG is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF.

Each side of Polygon FFF was multiplied by a certain value, known as the scale factor , to result in an area that is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF.

[Show me an example of how scale factor affects area]

\dfrac1{10}  

start fraction, 1, divided by, 10, end fraction

 

 

\begin{aligned} A &= \left(l\times\dfrac1{10}\right)\times\left(w\times\dfrac1{10}\right) \\ \\ A&= l\times w\times\dfrac1{10}\times\dfrac1{10} \\ \\ A&= lw \times \left(\dfrac1{10}\right)^2\end{aligned}  

 

 

 

 

 

 

 

 

\dfrac1{10}  

start fraction, 1, divided by, 10, end fraction\left(\dfrac1{10}\right)^2  

 

left parenthesis, start fraction, 1, divided by, 10, end fraction, right parenthesis, start superscript, 2, end superscript

Hint #22 / 3

The area of a polygon created with a scale factor of \dfrac1x  

x

1

​  start fraction, 1, divided by, x, end fraction has \left(\dfrac1{x}\right)^2(  

x

1

​  )  

2

left parenthesis, start fraction, 1, divided by, x, end fraction, right parenthesis, start superscript, 2, end superscript the area of the original polygon:

\left(\text{scale factor}\right)^2=\text{fraction of the area the scale copy has}(scale factor)  

2

=fraction of the area the scale copy hasleft parenthesis, s, c, a, l, e, space, f, a, c, t, o, r, right parenthesis, start superscript, 2, end superscript, equals, f, r, a, c, t, i, o, n, space, o, f, space, t, h, e, space, a, r, e, a, space, t, h, e, space, s, c, a, l, e, space, c, o, p, y, space, h, a, s

The area of Polygon GGG is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF. Let's substitute \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction into the equation to find the scale factor.

\left(\dfrac1{?}\right)^2=\dfrac19(  

?

1

​  )  

2

=  

9

1

​  left parenthesis, start fraction, 1, divided by, question mark, end fraction, right parenthesis, start superscript, 2, end superscript, equals, start fraction, 1, divided by, 9, end fraction

The scale factor is \dfrac13  

3

1

​  start fraction, 1, divided by, 3, end fraction.

Hint #33 / 3

Aimar used a scale factor of \dfrac13  

3

1

​  start fraction, 1, divided by, 3, end fraction to go from Polygon FFF to Polygon GGG.

4 0
3 years ago
Read 2 more answers
Se quiere construir un muro de 4 m de alto, 12 m de largo y 10 cm de espesor. ¿Cuántos ladrillos de 8 cm de alto, 20 cm de largo
Llana [10]

Answer:

3000

Step-by-step explanation:

Let's start by finding the volume of the wall. The volumen of the wall can be considered as the volume of a rectangular prism. The volume of a rectangular prism is given by:

V_w=w*l*h\\\\Where:\\\\w=Width=10cm=0.1m\\l=Length=12m\\h=Height=4m

So the volume of the wall is:

V_w=0.1*12*4=4.8m^3

Now, we can find the volume of the brick using the same method since a brick can be considered as a rectangular prism as well:

V_b=w*l*h\\\\For\hspace{3}the\hspace{3}brick\\\\w=10cm=0.1m\\l=20cm=0.2m\\h=8cm=0.08m

Hence:

V_b=(0.1)*(0.2)*(0.08)=0.0016m^3

In order to know how many bricks are required to build the wall, we just need to fill the wall volume with the number of bricks of this volume. So:

V_w=nV_b\\\\Where\\\\n=Number\hspace{3}of\hspace{3}bricks

Solving for n:

n=\frac{V_w}{V_b} =\frac{4.8}{0.0016} =3000

Therefore, we need 3000 bricks to build that wall.

Translation:

Comencemos por encontrar el volumen del muro. El volumen del muro puede considerarse como el volumen de un prisma rectangular. El volumen de un prisma rectangular viene dado por:

V_w=w*l*h\\\\Donde:\\\\w=Espesor=10cm=0.1m\\l=Largo=12m\\h=Alto=4m

Entonces el volumen del muro es:

V_w=0.1*12*4=4.8m^3

Ahora, podemos encontrar el volumen del ladrillo utilizando el mismo método, ya que un ladrillo también puede considerarse como un prisma rectangular:

V_b=w*l*h\\\\Para\hspace{3}el\hspace{3}ladrillo\\\\w=10cm=0.1m\\l=20cm=0.2m\\h=8cm=0.08m

Por lo tanto:

V_b=(0.1)*(0.2)*(0.08)=0.0016m^3

Para saber cuántos ladrillos se requieren para construir el muro, solo necesitamos llenar el volumen del muro con la cantidad de ladrillos de este volumen. Entonces:

V_w=nV_b\\\\Donde\\\\n=Numero\hspace{3}de\hspace{3}ladrillos

Resolviendo para n:

n=\frac{V_w}{V_b} =\frac{4.8}{0.0016} =3000

Por lo tanto, necesitamos 3000 ladrillos para construir ese muro.

4 0
3 years ago
QUESTION 20 PLEASE HELPPPP ASAPPPP
WARRIOR [948]

Answer: x=-1, y=-8

Step-by-step explanation: solve for the first variable in one of the equations then subsitutte the result into the other equation

4 0
3 years ago
ANOTHER ONE BESTIES< NO LINKS PLS!
zaharov [31]

Answer:

30 cards

Step-by-step explanation:

5 0
3 years ago
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