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The last three digits of $7^{2020}$ are $001$

Here is a link to a similar problem that should help you solve this: https://web2.0calc.com/questions/in-how-many-ways-can-you-spell-the-word-cool-in-the-grid-below-you-can-start-on-any-letter-then-on-each-step-you-can-step-one-letter-in-an

I dont think this requires a detailed explanation. The answer is 95.

We can start by expanding the equation (2x + 7)(x - 5) = -43 + jx ---> 2x^2 - (j+3)x + 8 = 0.

Now we know that when the discriminant (b^2 - 4ac) is equal to 0, we have one real solution.

So we plug in our values and we get j^2 + 6j - 55 = 0.

Solving this gives us j = -11, 5

We can start by expanding the equation (2x + 7)(x - 5) = -43 + jx ---> 2x^2 - (j+3)x + 8 = 0.

Now we know that when the discriminant (b^2 - 4ac) is equal to 0, we have one real solution.

So we plug in our values and we get j^2 + 6j - 55 = 0.

Solving this gives us j = -11, 5

This question has already been answered at https://web2.0calc.com/questions/help-please_11856