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3 years ago

7

Find and interpret the average rate of change.

1 answer:

miss Akunina [59]3 years ago

7 0

I will try my best friend

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Line segment AB has a length of 4 units. It is translated 2 units to the right on a coordinate plane to obtain line segment A'B'

Virty [35]

<span>length of the line segment does not change. you are just moving it 2 units to the right. answer would be 4 units</span>

6 0

3 years ago

Read 2 more answers

PLEASE HELP ME WITH THIS HELP ME WITH THIS MATH QUESTION IM NOT VERY GOOD AT THESE TYPE OF QUESTIONS (ONLY FIVE POINTS IM SORRY)

sergeinik [125]

1) 16% —> .16

.16 x $32,788 = $839.3728

2) 85/3 = 28.333333......

28.33....... - 20 = 8.333...... or rounded
to 8.2

3) 1 lb = .454 Kilos

15 lbs = ?

.454 x 15 = 6.81

15 lbs = 6.8 Kilos

8 0

4 years ago

A magician’s stage has a rectangle trapdoor which is (x+ 1/2) by 2x ft. The outside rectangle length is (x+ 16) ft.

Katen [24]

Answer:

Part A:- $x +11 ft$ Part B:- $(4x+54)ft$ part C:- $x= 2ft$

Part D:- Yes it satisfy his requirement

Step-by-step explanation:

Given that,

length and breadth of Trapdoor are $(x+ \frac{1}{2} ) by 2x$ ft.

The Outside length of Rectangle is $(x+16 )$ ft.

Part A:- Total area (in square feet) of the stage can be represented by

$x^{2} + 27x+ 176$ . Write an expression for the width of stage.

So, Area of rectangle = $length \times breadth$

$x^{2} + 27x+ 176 = (x+16) \times breadth$

$Breadth = \frac{x^{2} + 27x+ 176}{x+16}$

$Breadth = x+11$ ft. ................(1)

Part B:- Write an expression for the Perimeter of the stage.

Here, Perimeter of Rectangle = $2(L+B)$

= $2(x+16 + x +11)$

= $(4x + 54) ft$

Part C:- The area of trapdoor is $10 ft^{2}$ . Find the value of x.

So, Area of trapdoor = $10 ft^{2}$

$(x +\frac{1}{2}) \times 2x =10 ft^{2}$

$2x^{2} +x - 10 = 0$

$x = \frac{-1-9}{4}$ , $\frac{-1+9}{4}$

$x = \frac{-5}{2}$ , $2$

Hence value of x is $2 ft$ (Neglecting the negative value because

length cannot be negative).

Part D:- The magician wishes to have the area of the stage be at least 20 times the area of the trapdoor. Does this stage satisfy his requirement? Explain.

So, Length of trapdoor is $(2+\frac{1}{2} ) = \frac{5}{2} ft$

breadth of trapdoor is $2\times 2 = 4ft$

Now, outside length of Rectangle is = $18ft$

And outside breadth of rectangle is = $13ft$

Here, Area of trapdoor = $\frac{5}{2} \times 4 = 10ft^{2}$ ...................(2)

Area of rectangle = $18 \times 13= 234 ft^{2}$ ....................(3)

Thus comparing Equation (1) & (2) we found that

Area of rectangular stage is 20 times greater than the area of trapdoor.

7 0

4 years ago

♡♡♡ ▲ ▲ ▲ ▲ Ratio of hearts to triangles in simplest form.

Vitek1552 [10]

3 hearts

4 triangle

H:T

3:4

So, ratio of hearts to triangles in simplest form is 3:4

(we can't simplify any further - HCF of 3 & 4 is 1)

Hope this helps!

4 0

2 years ago

Question 1 (Multiple Choice Worth 5 points)

Kazeer [188]

It’s the last option if you need an explanation just let me know!

5 0

3 years ago