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dexar [7]
3 years ago
10

Can 3h and 11min be in the same unit

Mathematics
1 answer:
kaheart [24]3 years ago
7 0

Answer:

Yes.

Step-by-step explanation:

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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
3 years ago
Each month the fire department host a pancake feed. This month 75 people attended the pancake feed. This is 13 less than twice a
Kazeer [188]
75=2x-13
75+13=2x-13+13
88=2x
88÷2=2x÷2
44=x
7 0
3 years ago
Please help! Question posted above ^
emmainna [20.7K]

Answer:

( x^2 - 6 )^2

Step-by-step explanation:

  • x^4 -12x^2 +36
  1. Find one factor
  2. Factor by grouping
  • (x^2 - 6x) (x^2 - 6x)

*solution* ( x^2 - 6 )^2

8 0
3 years ago
I only need help on number 8 please
Setler [38]

Answer:

A y =

1

2

x − 6

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
The table shows some values of a function of the form y = ax2 + bx + c.
natali 33 [55]

Answer:

Value of constant term c is (-4)

Step-by-step explanation:

The given table represents a function which is in the form of a quadratic equation,

y = ax² + bx + c

We choose three points (3, -10), (4, -16) and (5, -24) from the table and satisfy the equation to get the values of a, b, and c.

For point (3, -10)

-10 = a(3)² + 3b + c

9a + 3b + c = -10 -------(1)

For point (4, -16)

-16 = a(4)² + 4b + c

16a + 4b + c = -16 ------(2)

For point (5, -24)

-24 = a(5)² + 5b + c

25a + 5b + c = -24 -----(3)

Equation (1) - equation (2)

(9a + 3b + c) - (16a + 4b + c) = -10 + 16

-7a - b = 6

7a + b = -6 ------(4)

Equation (2) - equation (3)

(16a + 4b + c) - (25a + 5b + c) = -16 + 24

-9a - b = 8

9a + b = -8 -------(5)

Equation (4) - Equation (5)

(7a + b) - (9a + b) = -6 + 8

-2a = 2

a = -1

From equation (4),

-7(1) + b = -6

b = -6 + 7

b = 1

From equation (1)

9(-1) + 3(1) + c = -10

-9 + 3 + c = -10

c = -10 + 6

c = -4

Therefore, the value of constant term c is (-4).

5 0
3 years ago
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