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2 years ago

What is the value of x?

Mathematics

1 answer:

Alexxx [7]2 years ago

5 0

Answer:

To find the value of x, bring the variable to the left side and bring all the remaining values to the right side. Simplify the values to find the result.

Step-by-step explanation:

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3 years ago

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statuscvo [17]

The question is incomplete without the diagram.

Answer:

Icosahedron + Tetrahedron

Octahedron +Tetrahedron

Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

Using all of the digits 1-9 (1, 2, 3, 4, 5, 6, 7, 8, 9) in that order and the operations +, - create 4 different equations with

alexandr1967 [171]

5 + 6 + 89

4 + 7 + 89

24 + 67 + 9

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7 0

4 years ago

Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

substitute in the formula

$d=\sqrt{(5-2)^{2}+(2-0)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

$d=\sqrt{(7-5)^{2}+(-1-2)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

$d=\sqrt{(5)^{2}+(-1)^{2}}$

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

8 0

4 years ago

Kindly answer this one. Thank you.

steposvetlana [31]

Answer:

See below.

Step-by-step explanation:

I'm assuming these questions are about the Midline Theorem (segment AL joins the midpoints of the non-parallel sides.

♦ The midline's length is the average of the lengths of the top and bottom parallel sides.

$AL=\frac{OR+CE}{2}$

Use this equation and substitute values given in each problem, then solve for the missing information.

1. AL = x, CE = 9, OR = 5

$x=\frac{9+5}{2}=7$

2. AL = <em>m</em> - 4, CE = 15, OR = 17

$m-4=\frac{15+17}{2}=16\\\\m-4=16\\\\\\m=20\\\\AL=20-4=16$

3. OR = y + 5, AL = 15, CE = 18

$15=\frac{(y+5)+18}{2}\\\\15=\frac{y+23}{2}\\\\30=y+23\\\\7=y\\\\OR = 7+5=12$

4 0

3 years ago