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3 years ago

Help me please really need it asap

Mathematics

2 answers:

DerKrebs [107]3 years ago

7 0

Answer:

A) 2/3

B) 4 sqrt 30/9 = y

C) P= 3/4 * sqrt 2

Step-by-step explanation:

A.) A = K*B

1= K * 3

K = 1/3

Now we will find the unknown A value

A = 1/3 * 2

A= 2/3

B) x=k* y^2

27 = k * 4^2

K= 27/16

Now , we will find the unknown value of y

10= 27/16 * y^2

10*16/27 = y^2

Sqrt of 160/27 =y

4 sqrt 10/3 sqrt 3 =y

4 sqrt 30/9 = y

C) p = k * sqrt Q

3= k * sqrt 16

3 = k * 4

K= 3/4

Now, when we substitute the value of k in to the variation

P= 3/4 * sqrt 2

matrenka [14]3 years ago

3 0

Answer:

Step-by-step explanation:

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5 <br> 2<br> +(12−7+x) if x=6

steposvetlana [31]

Answer:

the answer will be 6 or 10

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3 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

4 years ago

Sam collects baseball cards. He gave 6 cards to his brother. He then got 4 new cards from his father. Sam now has 12 cards. How

JulijaS [17]

Answer:

Step-by-step explanation:

If he had 14 then he gave away 6, he'd have 8. 8 plus 4 is 12. He had 14 before giving any away.

8 0

3 years ago

Read 2 more answers

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olga55 [171]

Answer:

A. Trapezoid

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3 years ago

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A street that is 360 feet long is covered in snow city workers are using a snow plow to clear the street a tire on the snow plow

belka [17]

Answer:

Radius of tire=0.58 meter

Step-by-step explanation:

as we know that the

Circumference of the tire=2*3.14*r

and it needs to rotate 30 times to cover the street distance of 360 feet

So, the below equation would be used to find the radius of tire in the meters

360/3.28084=30*2*3.14*r

r=0.5824203822 meters

Hence

$Radius=0.58 meter$

3 0

3 years ago

Help me please really need it asap​

Help me please really need it asap