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2 years ago

Can someone help me please

Mathematics

1 answer:

pshichka [43]2 years ago

7 0

Answer: 56 times

Pls mark brainleist i really need it

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14-12x+39x-18x=256-60x-657x

emmasim [6.3K]

Answer:

$\boxed{\bold{x=\frac{1}{3}}}$

Step-by-step explanation:

Combine Similar Elements: $\bold{-12x+39x-18x=9x}$

$\bold{14+9x=256-60x-657x}$

Combine Similar Elements: $\bold{-60x-657x=-717x}$

$\bold{14+9x=256-717x}$

Subtract 14 From Both Sides

$\bold{14+9x-14=256-717x-14}$

Simplify

$\bold{9x=-717x+242}$

Add $\bold{717x}$ To Both Sides

$\bold{9x+717x=-717x+242+717x}$

Simplify

$\bold{726x=242}$

Divide Both Sides By $\bold{726}$

$\bold{\frac{726x}{726}=\frac{242}{726}}$

Simplify

$\bold{x=\frac{1}{3}}$

4 0

3 years ago

Given O A ‾ ⊥ O C ‾ OA ⊥ OC start overline, O, A, end overline, \perp, start overline, O, C, end overline m ∠ B O C = 6 x − 6 ∘

Lelu [443]

Answer:

m∠BOC = 40°

Step-by-step explanation:

Given O A ‾ ⊥ O C ‾ OA ⊥ OC m∠BOC=6x−6 ∘

m∠AOB=5x+8 ∘

Find m ∠ B O C:

This means that: m∠BOC and m∠AOC intersect at a right angle.

Hence:

m∠BOC + m∠AOC = 90°

Step 1

Solving for x

6x - 6 + 5x + 8 = 90°

11x -2 = 90°

11x = 90 - 2

11x = 88

x = 88/11

x = 8

Step 2

Solving for m∠BOC

m∠BOC = 6x - 8

m∠BOC = 6(8) - 8

= 48 - 8

= 40°

7 0

3 years ago

Quadrilateral ABCD has the following vertices:

GenaCL600 [577]

9514 1404 393

Answer:

yes

Step-by-step explanation:

The figure can be shown to be a parallelogram by showing the sum of endpoints of the diagonals is the same.

A +C = B +D

(0, 6) +(0, -4) = (0, 2) = (3, 5) +(-3, -3) . . . . diagonals bisect each other

If the diagonals of a quadrilateral bisect each other, it is a parallelogram. A parallelogram with a right angle is a rectangle. So, ABCD is a rectangle.

_____

<em>Additional comment</em>

The midpoint of each diagonal is half the sum of the end point coordinates. That is, the midpoints are (0, 2)/2 = (0, 1). Since calculation of the midpoints requires both sums be divided by 2, we can tell the midpoints are the same if the sums are the same.