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3 years ago

12

Help me on my geometry hw it’s due in 25 minutes

2 answers:

icang [17]3 years ago

8 0

Answer:

now just half the first triangals answers

Step-by-step explanation:

x =21

y=24

<u>TH</u>=24 <u>PQ</u>=12

<u>RH</u>=24 <u>NQ</u>=12

<u>RT</u>=21 <u>NP</u>=10.5

>T=80 >P=80

>H=20 >Q=20

>R=80 >N=80

Paul [167]3 years ago

7 0

Answer:

x=10.5 y=12

Step-by-step explanation:

21/2=10.5

24/2=12

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A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

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Nataliya [291]

The customer should receive $6.37 back

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The total bill for the drinks and pizza comes out to be $17.28. How much should each person pay if they are to share the total b

sleet_krkn [62]

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Step-by-step explanation:

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2 years ago

Read 2 more answers

What does 6•7-27•9+64=

Natali [406]

Hi!

$6*7-27*9+64$

$42-27*9+64$

$42-243+64$

$Answer---->=-137$

Explanation: <u><em>This question is super super easy to me. This question is about how to use their order of operations the math problems. First you had to used PEMDAS stands for P-Parenthesis first, E-Exponents second, M-Multiply third, D-Divide fourth, A-Add fifth, and S-Subtract sixth. Then multiply and divide from left to right with 6*7=42. Then 27*9=243,42-243+64, and add or subtract from left to right with 42-243+64=-137 is the right answer. Hope this helps! And thank you for posting your question at here on Brainly. And have a great day. -Charlie</em></u>

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Solve the System of equations using substitution or elimination. Show work. 3x-4y=-31 X+y=6

Nataly_w [17]

Hope this helps I got (-1, 7)

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