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Valentin [98]
2 years ago
6

Which of the following values of y makes the equation 4y squared - 7y = 15 true?

Mathematics
1 answer:
uranmaximum [27]2 years ago
7 0

Answer:

C) y=3

Step-by-step explanation:

4y squared - 7y = 15

4(3) squared -7(3)=15

4(9) -7(3)=15

39-21=15

15=15

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Ms Smith has a monthly income of 4,500. If she spends $300 on clothing, and she spends 8% in entertainment , which expense is hi
lbvjy [14]

Answer:

Entertainment

Step-by-step explanation:

8% of $4,500 is $360, which is more than $300

8 0
3 years ago
What are the solution(s) of x^2-4=20?
Pie

x^2-4=20\qquad|+4\\\\x^2=24\to x=\pm\sqrt{24}\\\\x=-\sqrt{4\cdot6}\ \vee\ x=\sqrt{4\cdot6}\\\\x=-2\sqrt6\ \vee\ x=2\sqrt6

4 0
2 years ago
1/5 ÷ 2. simple for a couple of points lol
Sergio039 [100]

0.1

Have a amazing day!!

8 0
3 years ago
Read 2 more answers
Using principle of mathematical induction prove that 6^-1 divisble by 5 .​
salantis [7]

I suppose the claim is 5 \mid 6^n - 1 for n\in\Bbb N.

When n=1, we have 6^1 - 1 = 6 - 1 = 5, and of course 5 divides 5.

Assume the claim holds for n=k, that 5 \mid 6^k - 1. We want to use this to show it holds for n=k+1, that 5 \mid 6^{k+1} - 1.

We have

6^{k+1} - 1 = \left(6^{k+1} - 6\right) + \left(6 - 1) = 6\left(6^k - 1\right) + 5

Since 5 \mid 6^k - 1, we can write 6^k - 1 = 5\ell for some integer \ell. Then

6^{k+1} - 1 = 6\cdot5\ell + 5 = 5(6\ell + 1)

which is clearly divisible by 5. QED

5 0
1 year ago
What is the constant of variation for the quadratic variation? x 2 3 4 5 6 y 32 72 128 200 288
cluponka [151]
We use different models for different types of variation. For example, linear variation is associated with the formula y=ax, or the more familiar y=mx+b (the equation of a straight line). Cubic variation: y=a*x^3. In the present case we're discussing quadratic variation; perhaps that will ring a bell with you, reminding you that y=ax^2+bx+c is the general quadratic function. Now in y our math problem, we're told that this is a case of quadratic variation. Use the model y=a*x^2. For example, we know that if x=2, y =32. Mind substituting those two values into y=a*x^2 and solving for y? Then you could re-write y=a*x^2 substituting this value for a. Then check thisd value by substituting x=3, y=72, and see whether the resulting equation is true or not. If it is, your a value is correct. But overall I got 16!
7 0
3 years ago
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