Let the amount invested with 5% interest be x
Therefore, the amount invested with 6% interest will be (6000-x)
It is given that the total interest earned yearly is $337.5. Thus, the equation of interest will be:

Multiplying both sides by 100 we get:


Subtracting both sides by 36000 we get:


Thus, the amount invested at 5% is $2250
Therefore, the amount invested at 6% will be $(6000-2250)=$3750
Answer:
Option D
Step-by-step explanation:
In this question we use properties of parallelogram and angle sum property of a triangle.
In parallelogram ABCD

As, we know that opposite angles of parallelogram are equal
Therefore,

Now, in triangle ADC
We know that sum of all the angles of a triangle is 



Subtracting 97 from both sides we get


Measure of 
Answer: Phillip is correct. The triangles are <u>not </u>congruent.
How do we know this? Because triangle ABC has the 15 inch side between the two angles 50 and 60 degrees. The other triangle must have the same set up (just with different letters XYZ). This isn't the case. The 15 inch side for triangle XYZ is between the 50 and 70 degree angle.
This mismatch means we cannot use the "S" in the ASA or AAS simply because we don't have a proper corresponding pair of sides. If we knew AB, BC, XZ or YZ, then we might be able to use ASA or AAS.
At this point, there isn't enough information. So that means John and Mary are incorrect, leaving Phillip to be correct by default.
Note: Phillip may be wrong and the triangles could be congruent, but again, we don't have enough info. If there was an answer choice simply saying "there isn't enough info to say either if the triangles are congruent or not", then this would be the best answer. Unfortunately, it looks like this answer is missing. So what I bolded above is the next best thing.
Answer:
Step-by-step explanation:
11=4(8)+b
11=32+b
-21=b
y=4x-21
i think
Answer:
5.5193237 times 10 to the power of -10
Step-by-step explanation:
To solve you just need to divided the area by the length or width the question has provided. Please note I am in 7th grade.