All categories

2 years ago

Need some help on vertical adjacent complementary angles (algebraic)

Mathematics

2 answers:

dusya [7]2 years ago

6 0

Step-by-step explanation:

$(2c + 8) + (6c - 6) = 90 \\ 8c + 2 = 90 \\ 8c = 88 \\ c = \frac{88}{8} = 11$

sesenic [268]2 years ago

3 0

The figure will equal to 90 degrees.

So what you want to do is this:

1. Add them both, and equal

2. Add like terms,

and equal that

to 90 degrees.

3. Then minus 2

from both sides.

4. And you will 88 on the

right side, and 8c on the

left side.

5. Just divide from both sides.

2c+8+6c-6=90

8c+2=90

-2 -2

8c=88

/8 /8

c=11

You might be interested in

These two trapezoids are similar What is the correct way to complete the similarity statement?

pentagon [3]

Option A:

$\mathrm{ABCD} \sim \mathrm{GFHE}$

Solution:

ABCD and EGFH are two trapezoids.

To determine the correct way to tell the two trapezoids are similar.

Option A: $\mathrm{ABCD} \sim \mathrm{GFHE}$

AB = GF (side)

BC = FH (side)

CD = HE (side)

DA = EG (side)

So, $\mathrm{ABCD} \sim \mathrm{GFHE}$ is the correct way to complete the statement.

Option B: $\mathrm{ABCD} \sim \mathrm{EGFH}$

In the given image length of AB ≠ EG.

So, $\mathrm{ABCD} \sim \mathrm{EGFH}$ is the not the correct way to complete the statement.

Option C: $\mathrm{ABCD} \sim \mathrm{FHEG}$

In the given image length of AB ≠ FH.

So, $\mathrm{ABCD} \sim \mathrm{FHEG}$ is the not the correct way to complete the statement.

Option D: $\mathrm{ABCD} \sim \mathrm{HEGF}$

In the given image length of AB ≠ HE.

So, $\mathrm{ABCD} \sim \mathrm{HEGF}$ is the not the correct way to complete the statement.

Hence, $\mathrm{ABCD} \sim \mathrm{GFHE}$ is the correct way to complete the statement.

3 0

2 years ago

The school that molly goes to is selling tickets to a choral performance. On the first day of ticket sales the school sold 6 sen

krek1111 [17]

The cost of a senior citizen ticket is $15 and the cost of a student ticket is $12.

How did I get this?
We know that 6 citizen tickets and 7 student stickers sold for $174 the first day. And 10 citizen tickets and 14 student tickets sold for $318 the second day.

1. create two equations out of this: C= citizen cost per ticket and S = student cost per ticket.

6C + 7S = $174
10C + 14S = $318

2. Use process of elimination. Multiply the first equation by 2 because we want two variables to cancel out.

-12C - 14S = -$348
10C + 14S = $318
Combine like terms.

-2C = $30
Divide by -2 on both sides. The left side cancels out.

C = $30/-2
C = -$15 (In this case the negative doesn't matter)

C = $15 (cost of senior citizen ticket)

Plug the value of C into any of the two equations so we can get the value of S.

6($15) + 7S = $174
Distribute the 6 into the parenthesis.

$90 + 7S = $174
Subtract both sides by $90 and the left side will cancel out.

7S = $84
Divide both sides by 7.
S = $12

Student ticket: $12
Senior citizen ticket: $15

5 0

3 years ago

Which number line shows the solution set for |y-9| <12

RUDIKE [14]

Answer:

Option c

$y\geq-3$ or $y \leq 21$