0.25×m = 3.6 m=_____.

Answers

Answer 1
Answer: 0.25xm = 3.6m = m=0
Hope this answer helps

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Which situation can be represented by the
inequality 4x − 25 < 125?

Answers

First add 25 to both sides of the inequality.
You get 4x < 150
Now divide both sides by 4.
You get x < 37.5
So x can be anything less than 37.5,  such as 36,  35 or 34.

Combine the like terms to simplify the expression:9 + 3 + 17xy2 + 8y3 + 10x –13xy2 – 9x – 6y3
A. 2y6 + 4x2y2 + 10x + 12
B. 2y3 – 4xy2 – 3x + 12
C. 2y3 + 4xy2 + 10x + 9
D. 2y3 + 4xy2 + x + 12

Answers

9 + 3 + 17xy² + 8y³ + 10x - 13xy² - 9x - 6y³
rearrange like terms together
(in descending exponential order)
8y³ - 6y³ + 17xy² - 13xy² + 10x - 9x + 9 + 3
2y³ + 4xy² + x + 12

Answer is: D.2y³ + 4xy² + x + 12

What is the mid point for (16,-4)(-2-6)?

Answers

That would be (7, -5)

Midpoint Formula:
(x,y) = (x2+x1/2 , y1 + y2/2)
It would be (7,-5) there ya go

Calculate the perimeter of a rectangle with sides 9cm and 13cm

Answers

9 x 2= 18.   13x 2=26      26+18=44cm
So if 9cm +9cm=18cm then if you do 13cm +13cm=26cm then if you add 18cm + 26cm =44 cm

The scale of a square map indicates that each inch on the map corresponds to 5 miles. Write an expression that describes the area of land shown on the map. If the map is 8 inches on one side, what is the area of land shown on the map.

Answers

If x is the number of inches then the area is A = (5x)^2
So if x is 8 the area is (5*8)^2 = 1600 square miles

What is the distance, in feet, across the patch of swamp water?

Answers

Answer:

Therefore the distance across the patch of swamp water is 50 ft

Step-by-step explanation:

Given:

VW = 100 ft

WX = 60 ft

XZ = 30 ft

To Find:

ZY = l = ?

Solution:

In  Δ VWX and Δ YZX

∠W ≅ ∠ Z    …………..{measure of each angle is 90° given}

∠VXW ≅ ∠YXZ      ..............{vertically opposite angles  are equal}

Δ ABC ~ Δ DEC ….{Angle-Angle Similarity test}

If two triangles are similar then their sides are in proportion.

(VW)/(YZ) =(WX)/(ZX) =(VX)/(YX)\ \textrm{corresponding sides of similar triangles are in proportion}\n  

On substituting the given values we get

(100)/(l) =(60)/(30)\n\nl=(3000)/(60)=50\ ft

Therefore the distance across the patch of swamp water is 50 ft