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aliina [53]
4 years ago
13

Given: RHOM is a rhombus. Prove: MRH = MOH Need help for a friend

Mathematics
1 answer:
blsea [12.9K]4 years ago
6 0
A= given
b= congruent side
c= transitive property
d= def. of a rhombus
 
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Q # 13 please help me
Ede4ka [16]

It is false. Since you don't know what is the variable equal to. But multiplication can't be in associative property.

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3 years ago
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What kind of triangle is this!!! Please Help!!!
Arisa [49]

Answer:

Right Isosceles

Step-by-step explanation:

First, let's see if this triangle is Acute, Obtuse or Right.

We see that PB and QB are perpendicular lines, meaning that they form a right angle.  Therefore, triangle PQB is a right triangle.

Next, let's see if this triangle is equilateral, isosceles or scalene. PB and QB are congruent side lengths but PQ is not congruent to either PB or QB. Therefore, because two of the side lengths are congruent to each other and one is not, then triangle PQB is a isosceles triangle.

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3 years ago
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Answer: (0,4)

Step-by-step explanation:

3 0
3 years ago
Erica invests $10,000 at 5% interest compounded annually.
grigory [225]
A=P(1+i)^{n}
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8 0
3 years ago
Please help me with question 2
WARRIOR [948]

Answer:

sin(-a)= -15 /17

Step-by-step explanation:

Alright, lets remember that sin(x) is odd function, which means that

wherever you watch an expression like this: sin(-a), you can use the odd functions property, and turn it like the following

sin(-a)= - sin(a)

Now, other thing to remember: (sin(x))^{2}+(cos(x))^{2}=1

So, we can obtain any value for sin(x) function starting from the opposite function, cos(x), using the following:

sin(x)=\sqrt{1-(cos(x))^{2} }

So, if cos(a)= - 8 / 17, using the above equation we obtain the value for sin(a)

sin(a)=\sqrt{1 - (cos(a))^{2} } = \sqrt{1 - ( \frac{8}{17} )^{2}  }= \sqrt{\frac{225}{289} }  =15/17

Using the odd function property, we get the following result

sin(-a)= - sin(a) = -15/17

6 0
3 years ago
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