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2 years ago

Pls help i need this math question solved Brainleist to the first person

Mathematics

2 answers:

AleksAgata [21]2 years ago

7 0

Answer:

x = 10°

Step-by-step explanation:

8x + 50 = 130 (corresponding angles are equal)

8x = 80

x = 10°

Whitepunk [10]2 years ago

5 0

Answer:

x=58

Step-by-step explanation:

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The perimeter of a triangle is 7 m. The longest side is twice the length of the shortest side. The sum of the

34kurt

Answer:

See below in bold.

Step-by-step explanation:

Let the length of the shortest side be x, then the longest side = 2x.

Also x + y = 2x + 1 where y = length of the middle side,

Simplifying:

2x - x - y = -1

x - y = -1 (A)

Also as the perimeter = 7:

x + 2x + y = 7

3x + y = 7 (B)

Adding (A) and (B):

4x + 0 = 6

x = 6/4 = 1.5

Plug x = 1.5 into equation B:

3(1.5) + y = 7

y = 7 - 4.5 = 2.5.

Answer:

So the shortest side = 1.5, middle = 2.5 and longest = 2*1.5 = 3.

3 0

2 years ago

What is 9/10 divided by 3/5

Paha777 [63]

Answer:

1 /2 or 1.5

Step-by-step explanation:

Dividing fractions are relatively easy. One way to do it is multiplying the reciprocal of one of the fractions to the other. So 3/5 would become 5/3. Multiply, 9/10*5/3=45/30. Now that you have the answer you need to simplify which will end up into 1 1/2.

7 0

2 years ago

Read 2 more answers

PLEASE HELP THIS IS DUE BY TOMORROW

Bezzdna [24]

Answer:b

Step-by-step explanation:

4 0

2 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

2 years ago

The radius of a circle is 10 cm. Find its circumference in terms of pi

sergejj [24]

Answer:

20 pi is the correct answer

5 0

2 years ago