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ludmilkaskok [199]
3 years ago
13

Plesssse help me with this

Mathematics
1 answer:
Wewaii [24]3 years ago
6 0
I think it’s the fourth one
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Altitudes AA1 and BB1 are drawn in acute △ABC. Prove that A1C·BC=B1C·AC
Sophie [7]

Answer:

See the attached figure which represents the problem.

As shown, AA₁ and BB₁ are the altitudes in acute △ABC.

△AA₁C is a right triangle at A₁

So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)

△BB₁C is a right triangle at B₁

So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)

From (1) and  (2)

∴  A₁C/AC = B₁C/BC

using scissors method

∴ A₁C · BC = B₁C · AC

7 0
2 years ago
A sphere and a cylinder have the same radius and height. The volume of the cylinder is 11 ft.
Vedmedyk [2.9K]

Answer:

it 12

i litterally just got it right so...

3 0
2 years ago
The sides of a rectangle are in the ratio of 4:5. If the width is 28 in., find the length, the perimeter, and the area of this r
nadezda [96]
If the width is 28 inches, then divide that by 4 and you get 7. You multiply that by 5 to get the length. That would be 35. Just to check, you know that the width 28 and length 35 are in ratio 4:5 if you divide by 7. The perimeter would be 2(35+28)=63*2=126. So the perimeter is 126. The area would be 35*28 which is 980. To sum up, the answers are as follows.

Length: 35 in
Perimeter: 126 in
Area: 980 inches squared.
5 0
2 years ago
The sum of two numbers is less than 2. If we subtract the second number from the first, the difference is greater than 1. What a
Anna007 [38]

Answer:

the value of these two numbers are 1/2 and 3/2

Step-by-step explanation:

Given that:

x+y < 2 ---- (1)

y - x > 1 ---- (2)

From equation (2), let y  > 1 + x then substitute it into equation (1)

x + 1 + x < 2

2x + 1 < 2

2x < 2 - 1

2x < 1

x < 1/2

From equation (2), replace the value of x to be 1/2

y - 1/2 > 1

y > 1 + 1/2

y > 3/2

5 0
2 years ago
What is the value of the expression 5x − y when x = 3 and y = 1?<br><br> 14<br> 10<br> 7<br> 0
12345 [234]
5x-y
=5(3)-1
=15-1
=14

Therefore, the answer is 14.
5 0
3 years ago
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