Hey! Sorry that no one has answered your question in 4 days. But anyways, to do this, we need to simplify the problem a little. The mini marshmallows are unneeded, so we don't even have to think about them. Do do this, all you need to do is divide the 6 ounces of nuts and the 8 ounces of pretzels, or 6 / 8, to get 0.75.
Answer:
72
Step-by-step explanation:
1 then 7+1=8 8+7=15 15+8=23 23+15=38 38+23=61
Ta Da I have taught you the pattern now you can show your work just in case you have one of those teachers hope this helped
Answer: x = 9.6
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Explanation:
We have two smaller right triangles that are glued together so to speak.
The base of the smaller triangle on the left is 5 while the height is h.
Let's use the tangent rule to find the value of h
tan(angle) = opposite/adjacent
tan(55) = h/5
5*tan(55) = h
h = 5*tan(55)
h = 7.14074003371058
Make sure your calculator is in degree mode. That value of h above is approximate.
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Now focus on the smaller triangle on the right
It has the same height value h. This side is the adjacent side while x is the hypotenuse.
We'll use the cosine ratio
cos(angle) = adjacent/hypotenuse
cos(42) = h/x
cos(42) = 7.14074003371058/x
x*cos(42) = 7.14074003371058
x = 7.14074003371058/cos(42)
x = 9.6088135029715
x = 9.6
Answer:
<u>Part A: </u>
<u>n = 5q (1st equation)</u>
<u>0.05n + 0.25q = 2 (2nd equation)</u>
<u>Part B:</u>
<u>q = 4 ⇒ n = 5 * 4 = 20</u>
<u>Part C:</u>
<u>Margie has 20 nickels and 4 quarters, for a total of $ 2.00</u>
Step-by-step explanation:
Let's recall that a nickel has a value of $ 0.05 and a quarter a value of $ 0.25.
Let n represent the number of nickels and q represent the number of quarters.
Part A:
Write a system of equations to represent the situation.
n = 5q (1st equation)
n * 0.05 + q * 0.25 = 2
0.05n + 0.25q = 2 (2nd equation)
Part B:
Replacing n in the 2nd equation to solve for q:
0.05n + 0.25q = 2
0.05 * 5q + 0.25q = 2
0.25q + 0.25q = 2
0.5q = 2
q = 2/0.5
<u>q = 4 ⇒ n = 5 * 4 = 20</u>
<u>Part C:</u>
<u>Margie has 20 nickels and 4 quarters, for a total of $ 2.00</u>
