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2 years ago

Help I need a answer real quick! Plz don't send me a file type it out!!

Mathematics

2 answers:

melomori [17]2 years ago

7 0

The person above is correct:)

Andrews [41]2 years ago

4 0

Answer:

Step-by-step explanation:

2^2 = 4

√(4^2 + √5 ^2) = √21

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WILL MARK BRAINLIEST

ValentinkaMS [17]

Answer:

The answer to your question is:

x = 13

x + 10 = 23

5x - 42 = 23

Step-by-step explanation:

Data

angle 1 = x + 10

angle 2 = 5x - 42

They are vertical angles so, they measure the same.

x + 10 = 5x -42

x - 5x = -42 - 10

- 4x = -52

x = -52/-4

x = 13

angle 1 = x + 10 = 13 + 10 = 23

angle 2 = 5x - 42 = 5(13) -42 = 65 -42 = 23

8 0

2 years ago

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 $2a - 3b + 4c$

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

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

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

7 0

3 years ago

Two more than a number is less than 14. 2 + n less-than-or-equal-to 14 2 + n less-than 14 2 + n greater-than-or-equal-to 14 2 +

statuscvo [17]

Answer:

x+2 < 14 | -2

x < 12

2+n <= 14 | -2

n <= 12

2+n < 14 | -2

n < 14

2+n >= 14 | -2

n >= 12

2+n > 14 | -2

n > 14

8 0

2 years ago

Use the distributive property to remove the parentheses.

sergij07 [2.7K]

Answer:

3+2x

Step-by-step explanation:

Distribute the equation out.

1/4(12+8x)

(1/4*12)+(1/4*8x)

Multiple those out to get a new equation

3+2x

You cannot add those together because of the variable. Theres no way of knowing what x equals so this is as far as you can go therefor making the answer 3+2x

7 0

2 years ago

Read 2 more answers

Mya and ERIC WORK IN A BAKERY . MYA HAS 3 TIMEA AS MUCH FLOUR AS ERIC .MYA HAS 15.81LB OF FLOUR .USE THE EQUATION 3F = 15.81 TO

Pavel [41]

Answer:

5.27 lb

Step-by-step explanation:

Given that,

Mya has 3 times as much floor as Eric. Mya has 15.81 lb of floor. The relation is given by :

Let Eric has F floor.

3F = 15.81 ...(1)

We need to find how much flour Eric has.

Dividing both sides by 15.81.

F = 15.81/3

F = 5.27 lb

Hence, Eric has 5.27 lb of flour.

4 0

2 years ago