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mihalych1998 [28]

3 years ago

10

To increase the mean of 4 numbers by 2. by how much would the sum of the 4 numbers have to increase?

1 answer:

Orlov [11]3 years ago

4 0

So mean also translates to average so
the mean of 4 numbers is aprox
x=average
(x+x+x+x)/4=average
to increase x by 2 you add 2
x+2=average
(x+2+x+2+x+2+x+2)=4=(4x+8)/4=x+2

you would have to increase the sum/total by 8 to make the numbers' averages' go up by 2

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What is the dimensions of a box that is 11 lnches long 8 lnches wide and 6 in high

Sergio039 [100]

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Bad White [126]

(5) Angle ADC is the sum of angles ADB and BDC.

(6) Yes, because lines AG and EC form two pairs of vertical angles, including ADC and EDG, as well as ADE and CDG.

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3 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago

4. The equation of the line on the graph below is a direct variation equation. What is the constant of variation?

choli [55]

$\frac{1}{4}$ → a

the equation is y = mx ( where m is the slope / constant of variation )

calculate m using the gradient formula

m = ( y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁) = (0, 0 ) and (x₂, y₂ ) = (4, 1) ← 2 points on the line

m = $\frac{1-0}{4-0}$ = $\frac{1}{4}$

3 0

3 years ago