To solve this problem you must apply the proccedure shown below:
1- You have the following information given in the problem above:
- <span>The square of the diagonal of the rectangle is equal to the sum of the squares of the length and the width.
- The length is 25 meters and the diagonal is 45 meters.
2- Therefore, you have:
x: diagonal of the rectangle.
x^2=l^2+w^2
3- You have:
w^2+l^2-x^2=0
w^2+(25)^2-(45)^2=0
w^2-1400=0
w=37.41
The answer is: </span>w^2-1400=0
Step-by-step explanation:
= 55 - 2x is the answer .
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We need to find the remainder- the sticker left over
Divide 23 stickers into 4 piles = 23/4= 5 stickers per pile and 3 stickers left over
D. She would have 3 stickers left over
Answer:
72°
Step-by-step explanation:
From the information given:
A town planner wants to build two new streets, Elm Street and Garden Road, to connect parallel streets Maple Drive and Pine Avenue.
We are also told that there is a Trapezoid EFGH with EH as the Pine avenue and EF as the Elm street.
However, side FG and EH are parallel.
∠G = 108°
From the property of parallel lines :
since FG || EH
Then ∠G = ∠H = 108° (i.e corresponding angle will also be equal)
The required angle between Elm Street and Pine Avenue would be interior angles + 180° given that alternate angles are also equal.
The required angle between Elm Street and Pine Avenue = 180° - 108°
The required angle between Elm Street and Pine Avenue = 72°
Answer:
1600 integers
Step-by-step explanation:
Since we have a four digit number, there are four digit placements.
For the first digit, since there can either be a 5 or an 8, we have the arrangement as ²P₁ = 2 ways.
For the second digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.
For the third digit, since it neither be a 5 or an 8, we have two less digit from the total of ten digits which is 10 - 2 = 8. So, the number of ways of arranging that is ⁸P₁ = 8.
For the last digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.
So, the number of integers that can be formed are 2 × 10 × 8 × 10 = 20 × 80 = 1600 integers
