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Ugo [173]
3 years ago
8

a graphic designer created a logo on eight and a half by 11 inch paper. In order to be placed on a business card, the logo needs

to be one and 7/10 inches by 2 and 1/5 in. What is the scale factor of the dilation
Mathematics
1 answer:
agasfer [191]3 years ago
7 0

Answer:

The scale factor of the dilation is 1/5

Step-by-step explanation:


You might be interested in
Help me please (no links please)
stich3 [128]

a) This part is already complete I think..

b) This is a cuboid and lateral surface area of cuboid is: 2(lb +bh + hl)

= 2( 10 × 3 + 3 × 7 + 7 × 10)

= 2(30 + 21 + 70)

= 2 × 121 = 242 cm²

Now, the area of top & bottom: lb

= 2 × 10 × 3

= 60 cm²

Neglecting the top & bottom surface area of cuboid:

= 242 - 60

= 180 cm²

c) The total surface area us 242.. I have already done that part above...

__________________________

If i have done something wrong.. please lemme know :)

7 0
3 years ago
A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece
Alex_Xolod [135]

Answer:

The number of seat when n is odd S_n=\frac{n^2+2n+1}{4}

The number of seat when n is even S_n=\frac{n^2+2n}{4}

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

S_n=\frac{n}{2}[2a+(n-1)d]

    =\frac{n}{2}[a+l]

a = first term of the series.

d= common difference.

n= number of term

l= last term

n^{th} term of a A.P series is

T_n=a+(n-1)d

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=1 and d=2

n=1+(t-1)2

⇒(t-1)2=n-1

\Rightarrow t-1=\frac{n-1}{2}

\Rightarrow t = \frac{n-1}{2}+1

\Rightarrow t = \frac{n-1+2}{2}

\Rightarrow t = \frac{n+1}{2}

Last term l= n,, the number of term =\frac{ n+1}2, First term = 1

Total number of seat

S_n=\frac{\frac{n+1}{2}}{2}[1+n}]

    =\frac{{n+1}}{4}[1+n}]

     =\frac{(1+n)^2}{4}

    =\frac{n^2+2n+1}{4}

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=2 and d=2

n=2+(t-1)2

⇒(t-1)2=n-2

\Rightarrow t-1=\frac{n-2}{2}

\Rightarrow t = \frac{n-2}{2}+1

\Rightarrow t = \frac{n-2+2}{2}

\Rightarrow t = \frac{n}{2}

Last term l= n, the number of term =\frac n2, First term = 2

Total number of seat

S_n=\frac{\frac{n}{2}}{2}[2+n}]

    =\frac{{n}}{4}[2+n}]

     =\frac{n(2+n)}{4}

    =\frac{n^2+2n}{4}  

4 0
3 years ago
Which classification describes AMNO with vertices M(2, -3), N(3, 1), and<br> 0(-3, 1)?
Sonja [21]

Answer:

Option D

Step-by-step explanation:

32). Given vertices of the triangle are M(2, -3), N(3, 1) and O(-3. 1).

Distance between two points (x_1,y_1) and (x_2,y_2) is given by the expression,

Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Distance between M(2, -3) and N(3, 1) will be,

MN = \sqrt{(3-2)^2+(1+3)^2}

      = \sqrt{1+16}

      = \sqrt{17}

Distance between M(2, -3) and O(-3, 1),

MO = \sqrt{(2+3)^2+(-3-1)^2}

      = \sqrt{25+16}

      = \sqrt{41}

Distance between N(3, 1) and O(-3, 1),

NO = \sqrt{(3+3)^2+(1-1)^2}

      = 6

Condition for right triangle,

c² = a² + b² [Here c is the longest side of the triangle]

By this property,

MO² = MN² + NO²

(\sqrt{41})^2=(\sqrt{17})^2+6^2

41 = 17 + 36

41 = 51

False.

Therefore, given triangle is not a right triangle.

Since, length of all sides are not equal, given triangle will be a scalene triangle.

Option D is the correct option.

5 0
3 years ago
Use the Adding Integers manipulative to create a tile model for this situation: What is the sum of (-5) + (+3)?
Karo-lina-s [1.5K]

Answer:

-2

Step-by-step explanation:

(-5) + (+3)

= -5+3

= -2

please mark as brainliest

5 0
2 years ago
Read 2 more answers
How to solve 4x-3y=22,2x-y=10
Lady bird [3.3K]
Is that considered two equations?
4 0
3 years ago
Read 2 more answers
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