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4 years ago

5

50 POINTS!!!!! Name an angle that is complementary to ∠BOC.

2 answers:

Stolb23 [73]4 years ago

7 0

The complementary angle to ∠BOC is ∠COD.

BigorU [14]4 years ago

6 0

Angle COD a complementary angle is an angle that combines with another to create a 90 degree angle

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HELP<br> find the solution set of the inequality 15x-4 < 29.

seropon [69]

Answer: is actually 15×4 is 60 and 60 is greater than 29

Step-by-step explanation:

6 0

3 years ago

Idk how to do this honestly

Anuta_ua [19.1K]

I get the answer being the same y= -1/2x +

Because b in intercept form is "0"

I used
m=y2-y1/x2-x1
M=-3-2/6-4
M=-5/10
M=-1/2

(-4,2)
Y=mx+b
2=-1/2 (-4)+b
B=2-(-1/2)(-4)
B=0

I did the same for second point
And got "0" for b

So my answer is get
Y=-1/2x

Unless someone else gets something else different.

I hope somewhat helps

8 0

3 years ago

Giving brainliest!!!! please help!!

liberstina [14]

Answer:

{4, 5 }

Step-by-step explanation:

Solve the inequality

n + 1 > 4 ( subtract 1 from both sides )

n > 3

The set of values that make the inequality true is { 4, 5 }

8 0

3 years ago

1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera

Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

$SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}$

$SP \approx 8.246$

$PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}$

$PA \approx 8.246$

$AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}$

$AZ \approx 8.246$

$ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}$

$ZS \approx 8.246$

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

$m_{SA} = \frac{6-(-4)}{6-(-4)}$

$m_{SA} = 1$

$m_{PZ} = \frac{4-(-2)}{-2-4}$

$m_{PZ} = -1$

Since $m_{SA}\cdot m_{PZ} = -1$ , diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

$M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)$

$M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)$

$M_{SA} = (-2,-2)+(3,3)$

$M_{SA} = (1,1)$

$M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)$

$M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)$

$M_{PZ} = (2,-1)+(-1,2)$

$M_{PZ} = (1,1)$

Then, the diagonals SA and PZ bisect each other.

8 0

3 years ago

A baker bakes 752 muffins in a weekend. she sells the muffins in packs of 12. how many packs of muffins did she sell, how many d

eduard

752/12 = 62.66 and since she can't sell a portion of a 12 pack, she sold 62 whole packages and had 2/3 of one remaining, 2/3 of 12 is 8 remaining.

8 0

4 years ago