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Airida [17]
3 years ago
14

Which situation could NOT represent a proportional relationship?

Mathematics
1 answer:
My name is Ann [436]3 years ago
5 0

Answer:

The weight in <em>w</em> weeks of a baby that gains 1.5 pounds per month if her starting weight is 7.5 pounds.

Step-by-step explanation:

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Find the sum that amounts to rs.3025 at 10% p.A
Brrunno [24]

Answer:

P = Rs2500

Step-by-step explanation:

Given that,

Amount, A = Rs 3025

Rate, r = 10%

Let time, t = 2 years

We need to find the sum if the amount is compounded annually. Using the formula of compound interest.

A=P(1+\dfrac{r}{100})^n\\\\P(1+\dfrac{10}{100})^2=3025 \\\\P(1.21)=3025\\\\P=\dfrac{3025}{1.21}\\\\P=\text{Rs}\ 2500

So, the sum of money is Rs 2500.

7 0
2 years ago
Paul predicts that n stalks of corn planted together will produce (2n2–2n)(2n2–2n) ears of corn. If Paul plants three corn plant
Free_Kalibri [48]
Given that the production of n stalks of corn planted has been modeled by (2n^2-2n), then the approximate ears of corn produced by 3 stalks will be given by:
2n^2-2n
plugging in n=3 in the formula we get:
2(3)^2-2(3)
simplifying we get:
2(9)-6
=18-6
=12
the answer is: 12 ears
6 0
3 years ago
CAN SOMEONE HELP PLEASE <br> m4A=42°, b=10, c= 12. What is the length of a to the nearest tenth?
kipiarov [429]

<u>Answer:</u>

a = 8.1

<u>Step-by-step explanation:</u>

Use the cosine rule:

a = \sqrt{b^2 + c^2 -2 bc \space\ cosA }

Substituting the values:

a = \sqrt{10^2 + 12^2 -2(10)(12) \space\ cos 42 \textdegree\ }

⇒ a = \sqrt{4 \space\ cos 42\textdegree\ }

⇒ a = 8.1

4 0
2 years ago
1. x=1/(root3-root2). find rootx-(1/rootx) 2. if x=[root(a+2b)+root(a-2b)]/[root(a+2b)-root(a-2b]. show that bx^2-ax+b=0
timofeeve [1]
1. x = 1/ ( \sqrt{3} -  \sqrt{2)} = \sqrt{3}+  \sqrt{2};
 ( \sqrt{x} -1/ \sqrt{x} )^{2} = x + 1/x - 2 = = 
6 0
3 years ago
Read 2 more answers
I need help trying to do this
GREYUIT [131]

Answer:

me likey points ❤️❤️❤️

5 0
3 years ago
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