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3 years ago

I need the answer. (will give brainliest)

Mathematics

2 answers:

raketka [301]3 years ago

4 0

Answer:

the ans is 6

Step-by-step explanation:

bcz if u divide 15 by 3 =5 and 12by 3 =4 &18 by 3 =6

MatroZZZ [7]3 years ago

3 0

Answer:

Step-by-step explanation:

We can use ratios to solve

15 ft 18ft

---------- = -----------

5ft x

Using cross products

15x = 5*18

15x = 90

Divide by 15

15x/15 = 90/15

x =6

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A circle has a radius of 15 inches. What is the area, in square inches, of this circle? (The area of a circle with a radius of r

kozerog [31]

Answer:13

Step-by-step explanation:

7 0

4 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

I need help with these problems please

katovenus [111]

Answer:

1. Reflection

2. Dilation

3. Rotation

Step-by-step explanation:

Hey!

You can tell number one is a reflection because it didn’t change side or turn upside down or anything, it just flipped over to the other side.

You can tell number two is a dilation because it changes size.

You can tell number three is rotation because it rotated.

Hope this helps! :D

If you have any further questions, comment down below!

8 0

3 years ago

Triangle DEF has vertices D(0,0), E(7,0), and F(3,7).

Leni [432]

Answer:

(3, 1.7)

Step-by-step explanation:

The point at which the vertices of a triangle meet is known as the orthocenter of the triangle. The orthocenter passes through the vertex of the triangle and is perpendicular to the opposite sides.

Two lines are perpendicular if the product of their slopes is -1.

The slope of the line joining D(0,0), F(3,7) is:

$m_1=\frac{7-0}{3-0}=\frac{7}{3}$

The slope of the line perpendicular to the line joining D and F is -3/7. The orthocenter is perpendicular to the line joining D and F and passes through vertex E(7, 0). The equation is hence:

$y-y_1=m(x-x_1)\\\\y-0=-\frac{3}{7} (x-7)\\\\y=-\frac{3}{7}x+3 \ .\ .\ .\ (1)$

The slope of the line joining E(7,0), and F(3,7). is:

$m_1=\frac{7-0}{3-7}=-\frac{7}{4}$

The slope of the line perpendicular to the line joining E and F is 4/7. The orthocenter is perpendicular to the line joining E and F and passes through vertex D(0, 0). The equation is hence:

$y-y_1=m(x-x_1)\\\\y-0=\frac{4}{7} (x-0)\\\\y=\frac{4}{7} x\ .\ .\ .\ (2)$

The point of intersection of equation 1 and equation 2 is the orthocenter. Solving equation 1 and 2 simultaneously gives:

x = 3, y = 1.7

4 0

3 years ago

three bells ring at the intervals of 10 minutes and 20 minutes if they all ring together at 10 clock when will they ring togethe

mario62 [17]

Answer:The LCM of 10, 15, and 20 is 60. After 60 minutes, they ring together again.

Step-by-step explanation:

6 0

3 years ago