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AnnyKZ [126]
3 years ago
11

HELP PLEASE 98 points

Mathematics
1 answer:
castortr0y [4]3 years ago
7 0

Answer:

Given: ​ BD ​is an altitude of △ABC .

Prove: sinA/a=sinC/c

Triangle ABC with an altitude BD where D is on side AC. Side AC is also labeled as small b. Side AB is also labeled as small c. Side BC is also labeled as small a. Altitude BD is labeled as small h.

Statement Reason

​ BD is an altitude of △ABC .

Given △ABD ​ and ​ △CBD ​ are right triangles. (Definition of right triangle)

sinA=h/c and sinC=h/a

Cross multiplying, we have

csinA=h and asinC=h

(If a=b and a=c, then b=c)

csinA=asinC

csinA/ac=asinC/ac (Division Property of Equality)

sinA/a=sinC/c

This rule is known as the Sine Rule.

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Answer:

False

Step-by-step explanation:

The Pythagorean theorem can only be applied to right triangles.

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3/4 divided by 6/1 <br> How ?
AfilCa [17]
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Keep Flip Change
Keep the first fraction
3/4
Flip the second fraction(reciprocal)
6/1 -> 1/6
Change the sign
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So
3/4 × 1/6
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Someone help me please!<br><br> Solve the equation 1/4t + -6 = -4
Rom4ik [11]

Answer:

The answer is t=8

Step-by-step explanation:

simplify both sides or the equation, then isolate the variables

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For a school fundraiser, Elena will make T-shirts. She will order T-shirts and print graphics on them. Elena must spend $350 on
daser333 [38]

Answer:

Elena can make 150 t-shirts

Step-by-step explanation:

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7 0
3 years ago
Please help me solve this problem
Karo-lina-s [1.5K]

3.1 a) x = 6 / 5 or x = 2, b) x = - 3 / 2 or x = - 1 / 4, c) x = - 3 or x = 2, d) x = 1 or x = - 2.

3.2 a) x = 9 or x = - 3, b) x = 1 or x = - 1 / 4, c) x = - 3 or x = - 5 d) x = 8 or x = - 18

<h3>How to solve algebraic problems by appying absolute value properties</h3>

In this question we have eight expressions involving <em>absolute value</em> expressions, which can be solved by using the following procedure:

3.1 a) |(1 / 2) · x| = 3 - 2 · x

For x ≥ 0:

(1 / 2) · x = 3 - 2 · x

(5 / 2) · x = 3

x = 6 / 5

For x < 0:

- (1 / 2) · x = 3 - 2 · x

(3 / 2) · x = 3

x = 2

b) |x - 1| = 3 · x + 2

For x ≥ 1:

x - 1 = 3 · x + 2

2 · x = - 3

x = - 3 / 2

For x < 1:

- x + 1 = 3 · x + 2

4 · x = - 1

x = - 1 / 4

c) |5 · x| = x - 12

For x ≥ 0:

5 · x = x - 12

4 · x = - 12

x = - 3

For x < 0:

- 5 · x = x - 12

6 · x = 12

x = 2

d) |7 - x| = 5 · x + 1

For x ≤ 7:

7 - x = 5 · x + 1

6 · x = 6

x = 1

For x > 7:

x - 7 = 5 · x + 1

4 · x = - 8

x = - 2

3.2 a) |9 + x| = 2 · x

For x ≥ - 9:

9 + x = 2 · x

x = 9

For x < 9:

- 9 - x = 2 · x

3 · x = - 9

x = - 3

b) |5 · x| - 3 · x = 2

For x ≥ 0:

|5 · x| = 2 + 3 · x

5 · x = 2 + 3 · x

2 · x = 2

x = 1

For x < 0:

- 5 · x = 2 + 3 · x

- 8 · x = 2

x = - 1 / 4

c) |x + 6| - 9 = 2 · x

For x ≥ - 6:

x + 6 - 9 = 2 · x

x - 3 = 2 · x

x = - 3

For x < - 6:

- x - 6 - 9 = 2  · x

- x - 15 = 2 · x

3 · x = - 15

x = - 5

d) |2 · x - 3| + x = 21

For x ≥ 3 / 2:

2 · x - 3 + x = 21

3 · x = 24

x = 8

For x < 3 / 2:

- 2 · x + 3 + x = 21

- x = 18

x = - 18

To learn more on absolute values: brainly.com/question/1301718

#SPJ1

3 0
2 years ago
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