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3 years ago

5

laurie will draw a scale model of the garden she wants to plant. Her scale will be 1cm = 2.5 ft. what will be the actual dimensi

ons of laurie's? a. 1.6 ft by 3.4 ft b. 4 ft by 34 ft c. 8 ft by 34 ft d. 10 ft by 21.25 ft

2 answers:

g100num [7]3 years ago

4 0

Answer:

D

Step-by-step explanation:

Because i said.......

uranmaximum [27]3 years ago

4 0

The answer would be D. 4 x 2.5 is 10, and 8.5 x 2.5 is 21.25. Remember that 1 cm is equal to 2.5 feet.

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A local hamburger shop sold a combined total of 754 hamburgers and cheeseburgers on Saturday. There was 54 more cheeseburgers so

DochEvi [55]

Answer: 323 hamburgers

754/2=377
377 ham 377 cheese
54 more cheese than ham
377 ham - 54 = 323 hamburgers total sold on Saturday

7 0

2 years ago

I need help. Thank you.

Oduvanchick [21]

Answer:

D

A=-5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A transformer increases the input voltage from 12V to 12,000V. What is the ratio of the number of wire turns on the primary coil

erma4kov [3.2K]

Answer:

$\frac{n_p}{n_s}=\frac{1}{1000}$

Step-by-step explanation:

Given voltage of primary coil is $V_p=12V$

And voltage of secondary coil is $V_s=12,000V$

A transformer is a device that contains primary and secondary coil. These coils are mounted on soft iron. The change in voltage depends upon the number of turns of primary coil $(n_p)$ and secondary coil $(n_s)$ .

Following equation represents the relation between them.

$\frac{V_s}{V_p} =\frac{n_s}{n_p}\\\\\frac{12,000}{12}=\frac{n_s}{n_p}\\\\1000=\frac{n_s}{n_p}$

Then, ratio of the number of wire turns on the primary coil to the number of turns on the secondary coil is

$\frac{n_p}{n_s}=\frac{1}{1000}$

5 0

3 years ago

Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt

Leni [432]

Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = $\sqrt{Δx^2+Δy^2}$
d₁ = $\sqrt{(-2-1)^2+(1-4)^2}$
d₂ = $\sqrt{(x-1)^2+(y-4)^2}$
d₁ = d₂
∴ $\sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2}$ ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = $\pm \sqrt{9} = \pm 3$
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1) or (-2,7)

8 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