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3 years ago

HELP ASAP ASAP HELPPP

Mathematics

1 answer:

telo118 [61]3 years ago

3 0

I don’t think this would exist because it would be only on the -2 point.

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MakcuM [25]

Are these all different questions?

6 0

3 years ago

Melinda can paint 1/4 of a wall in the same time that Desiree can paint 2/5 of a wall. How much of a wall will Melinda have pain

Harlamova29_29 [7]

Answer:

Melinda will paint 5/8 part of a wall when Desiree has finished painting 1 wall.

Step-by-step explanation:

It is given that:

Melinda can paint 1/4 of a wall in the same time that Desiree can paint 2/5 of a wall.

Let 't' be the time taken by Melinda and Desiree to paint 1/4 and 2/5 of a wall respectively.

Now the time taken by Desiree to paint 1 wall will be:

2/5 of wall= t

1 wall= (5/2) t

Now in this time ( i.e. 5/2 t) the part of a wall painted by Melinda will be:

As Melinda takes-- t time to paint= 1/4 of a wall

5/2 t she will paint= (5/2)×(1/4) of a wall= (5/8) of a wall.

Hence Melinda will paint 5/8 part of a wall when Desiree has finished painting 1 wall.

3 0

3 years ago

What is the domain of this and how :-<br>f(x)=tanx

Dvinal [7]

Step-by-step explanation:

<u>So the domain of f(x) := tanx is all real numbers except x = π 2 + kπ, k an integer.</u>

$\huge\bold\red {Befikra❥☘⍣❦}$

3 0

2 years ago

The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2

Arada [10]

Given:

Scale factor $s=\dfrac{1}{3}$

Center of dilation = (4,2)

To find:

The coordinates of the points C' and A.

Solution:

We know that, if a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

$(x,y)\to (k(x-a)+a,k(y-b)+b)$

The scale factor is $\dfrac{1}{3}$ and the center of dilation is at (4,2).

$(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)$ ...(i)

Suppose the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using rule (i), we get

$C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)$

$C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)$

$C(-2,11)\to C'(-2+4,3+2)$

$C(-2,11)\to C'(2,5)$

Hence, the coordinates of Point C' are C'(2,5).

Let us assume that point A is A(m,n).

Using rule (i), we get

$A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)$

From the given figure it is clear that the image of point A is (8,4).

$A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)$

On comparing both sides, we get

$\dfrac{1}{3}(m-4)+4=8$

$\dfrac{1}{3}(m-4)=8-4$

$(m-4)=3(4)$

$m=12+4$

$m=16$

And,

$\dfrac{1}{3}(n-2)+2=4$

$\dfrac{1}{3}(n-2)=4-2$

$(n-2)=3(2)$

$n=6+2$

$n=8$

Therefore, the coordinates of point A are (16,8).

4 0

3 years ago

I need some help plz!!!

jek_recluse [69]

1.04 kilograms

1 gram = .001 kilogram
2x520= 1040
1040 x .001 = 1.04 kilograms

5 0

3 years ago