All categories

3 years ago

Help please ?????? I don't understand

Mathematics

1 answer:

Stells [14]3 years ago

4 0

Answer:

It's going up by a difference of 70, & 2.

Step-by-step explanation:

You might be interested in

Describe how to find the scale factor if an antenna is 60 feet long and a scale drawing shows the length as 1 foot long. Please

bekas [8.4K]

The actual is 60 feet long, the drawing is only 1 foot long. So the scale factor must be 1/60. You know that the scale factor has to be less than one in this case because you're going from a bigger measurement to a smaller measurement(60 to 1).

3 0

2 years ago

Help please brainliest

serg [7]

Answer:

d.4x - 6

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

What is the fraction for What is the fraction for 0.21

gavmur [86]

The fraction for 0.21 would be 21/100

3 0

2 years ago

12. Find the surface area of each figure to the nearest tenth.

Zolol [24]

Answer:

12a. 471.2 cm²

12b. 60 m²

Step-by-step explanation:

Part A.

The surface area of each figure is the sum of the end area and the lateral area.

cylinder

S = (2)(πr²) +2πrh = 2πr(r +h)

S = 2π(5 cm)(5 cm +10 cm) =150π cm² ≈ 471.2 cm²

triangular prism

S = (2)(1/2)bh + PL . . . . b=triangle base; h=triangle height; P=triangle perimeter; L=length of prism

S = (4 m)(1.5 m) + ((4 + 2·2.5) m)(6 m) = (6 + 54) m² = 60 m²

_____

Part B.

Surface area is useful in the real world wherever products are made from sheets of material or wherever coverings are applied.

Carpeting or other flooring, paint, wallpaper are all priced in terms of the area they cover, for example.

The amount of material used to make containers in the shapes shown will depend on the area of these containers (and any material required for seams).

6 0

2 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.

7 0

3 years ago