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DaniilM [7]
3 years ago
6

PLEASE HELP MATH MADD POINTS

Mathematics
2 answers:
igomit [66]3 years ago
7 0
It should be B. 
The rate of change is the same as the slope.
timama [110]3 years ago
3 0

Answer:

the slope of each trend line.

Step-by-step explanation:

Just took the test for edge.

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PLEASE HELP IM ON A TIME LIMIT :(
Mice21 [21]

Answer:

hey hope this helps

<h3 /><h3>Comparing sides AB and DE </h3>

AB =

\sqrt{ {1}^{2} +  {1}^{2}  }

=  \sqrt{2}

DE

= \sqrt{ {(3 - 5)}^{2} +  {(1 + 1}^{2}  }  \\   = \sqrt{ {( - 2)}^{2}  +  {(2)}^{2} }  \\    = \sqrt{4 + 4}  \\  =  \sqrt{8}  \\   = 2 \sqrt{2}

So DE = 2 × AB

and since the new triangle formed is similar to the original one, their side ratio will be same for all sides.

<u>scale factor</u> = AB/DE

= 2

It's been reflected across the Y-axis

<em>moved thru the translation of 3 units towards the right of positive x- axis </em>

for this let's compare the location of points B and D

For both the y coordinate is same while the x coordinate of B is 0 and that of D is 3

so the triangle has been shifted by 3 units across the positive x axis

7 0
3 years ago
How many packs of pencils can you get if there is 1,440 pencils and 6 in each pack
mamaluj [8]

Divide 1440 by 6 to get 240 packs.

4 0
3 years ago
Find the equation of a line that passes through points (1,8) and (4,4).​
daser333 [38]

Step-by-step explanation: Using the equation, y = mx + b, the slope-intercept form, plug in the values you have and solve for b. (m is the slope.) You have y = -8 and x = 4 and m = 1/2.

3 0
3 years ago
1. Pia printed two maps of a walking trail. The length of the trail on the first map is 8 cm. The length of the trail on the sec
Alex

Answer:

Step-by-step explanation:

b) 1- scale factor from the first map to the second map:

\frac{8}{6}  = 1.33

   2- landmark on the first map is a triangle with side lengths of 3 mm, 4 mm, and 5 mm.

Side lengths of the landmark on the second map

Divide the length by scale factor:

side lengths of 3 mm:  \frac{3}{1.33} = 2.25 mm

side lengths of 4 mm:  \frac{4}{1.33} = 3.007 mm

side lengths of 5 mm:  \frac{5}{1.33} = 3.75 mm

5 0
3 years ago
The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC
DIA [1.3K]

Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x²  ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)²  ⇒(2)

Equating (1) and (2)

∴  10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴  10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²

∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

∴ ΔPAQ is a right triangle at A,

PA = 2sqrt5 in  and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

6 0
3 years ago
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