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Yanka [14]
3 years ago
15

HEEELLLLPPPPP PLEASE

Mathematics
1 answer:
Bond [772]3 years ago
5 0
What do you need help with?
You might be interested in
What is the value of this please help
mars1129 [50]
The symbol  √  means "square root of".   √16  means "square root of 16".
It's a number that gives  16  when you multiply the number by itself.

There are <u>two</u> numbers that can both do the job.    They are +4 and -4.

(+4) times (+4) = 16

(-4) times (-4) = 16
6 0
4 years ago
Help plz..And No links!! I repeat No links!!
Snowcat [4.5K]

Step-by-step explanation:

By Basic Proportionality Theorem, the triangles UTV and USR are similar.

Let us consider RV as x.

\tt{\dfrac{39}{26} = \dfrac{33}{x}}

\rm{x = \dfrac{33 \times 26}{39}}

\sf{ x = 22 } is the required answer, after simplifying it.

------------

Or if you want to prove it similar:-

In triangle UTV and USR,

Angle TUV = Angle SUR (common)

Angle UVT = Angle URS (Corresponding angles b/w parallel lines)

By AA similarity,

UTV =~ USR.

Hope it helps :)

4 0
3 years ago
PLS HURRY!!!<br> Solve the equation. What is the value of x <br><br> 1/3 (x + 1) + 2x=2
Dafna1 [17]

Answer:

Decimal: -0.5

Fraction: - 1/2

Step-by-step explanation:

7 0
3 years ago
Standard form 6000000+80000+10
SIZIF [17.4K]
6,080,010 hope this helps you 

8 0
4 years ago
Read 2 more answers
A 22m ladder and a 20m ladder were leaned against a building. The bottom of the longer ladder was 4m farther from the building t
OLga [1]

Answer:

Both ladder reaches 18.1 m up the building ⇒ 3rd answer

Step-by-step explanation:

* Lets study the information to solve the problem

- There are two ladders

- The lengths of them are 22 m and 20 m

- The bottom of the longer was 4 m farther than the bottom of the

  shorter from the building

- Both of them reached the same distance up the building

* Lets solve the problem

- Let the distance between the bottom of the shorter ladder to the

 building is x

∵ The bottom of the longer ladder is farther by 4

∴ The distance between the bottom of the longer ladder and the

   building is x + 4

- Let the ladders reached the distance h up the building

* Now we have two right triangles

# Their hypotenuses are 22 and 20

# Their heights are h

# Their bases are x + 4 , x

- Lets find h in each triangle using the rule of Pythagoras

∵ (hypotenuse)² = (leg 1)² + (leg 2)²

# The longer ladder

∵ hypotenuse = 22

∵ leg 1 = x + 4

∵ leg 2 = h

∴ (22)² = (x + 4)² + h² ⇒ simplify

∴ 484 = (x + 4)² + h² ⇒ subtract (x + 4)² from both sides

∴ h² = 484 - (x + 4)² ⇒ (1)

# The shorter ladder

∵ hypotenuse = 20

∵ leg 1 = x

∵ leg 2 = h

∴ (20)² = (x )² + h² ⇒ simplify

∴ 400 = x² + h² ⇒ subtract x² from both sides

∴ h² = 400 - x² ⇒ (2)

- Equate (1) , (2) to find x

∴ 484 - (x + 4)² = 400 - x² ⇒ Add (x + 4)² and subtract 400 in both sides

∴ 84 = (x + 4)² - x² ⇒ open the bracket

∴ 84 = x² + 2(4)(x) + 4² - x² ⇒ simplify

∴ 84 = 8x + 14 ⇒ subtract 16 from both sides

∴ 68 = 8x ⇒ divide both sides by 8

∴ x = 8.5

- Substitute this value of x in (1) or (2) to find h

∵ h² = 400 - x²

∴ h² = 400 - (8.5)² = 327.75 ⇒ take √ for both sides

∴ h = 18.1

* Both ladder reaches 18.1 m up the building

4 0
3 years ago
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