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Ilia_Sergeevich [38]
2 years ago
14

I need help and the answer for this practice question please help

Mathematics
1 answer:
muminat2 years ago
4 0

Answer:

The answer is ∆XWY. The reason being both sides are congruent.

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Rewrite the expressions in each pair so that they have the same base. c) (1 /2)^2x and (1/ 4)^x - 1
just olya [345]
I'm not completely sure but this is what I would do.
evaluate <span>(1/ 4)^x - 1 </span>as is. But change the (1 /2)^2x to (2/4)^2x. This way both fractions have the same denominator and in this sense, the same base. The 2/4 base still evaluates into 1/2 so nothing, mathematically, is being broken here.
7 0
4 years ago
Find a and k so that the given points lie on the parabola (An algebraic solution is required)
Aleks [24]

For the given points to lie on the parabola,

a = -3 and k = 10.

An equation of a curve that has a point on it that is equally spaced from a fixed point and a fixed line is referred to as a parabola. The parabola's fixed line and fixed point are together referred to as the directrix and focus, respectively. It's also crucial to remember that the fixed point is not located on the fixed line. A parabola is a locus of any point that is equally distant from a given point (focus) and a certain line (directrix).

According to the question,

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Points A(1,7) and B(4,-2)

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7 = a + k

Similarly,

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Learn more about parabolas here:

brainly.com/question/4061870

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4 0
2 years ago
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Harrizon [31]

Answer:

  44.2 years

Step-by-step explanation:

If we assume the interest is compounded annually and the investment is a one-time deposit into the account, its value each year is multiplied by 1+6.25% = 1.0625. After n years, the value in the account will be ...

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Dividing by 1300 and taking logs, we have ...

  log(19000/1300) = n·log(1.0625)

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It will take about 44.2 years for the account to reach $19,000.

5 0
3 years ago
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The pink paper for the book Giver
Leni [432]
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4 0
3 years ago
Plz help me with this <br> Answer options: Quadratic, exponential, cubic and linear
ElenaW [278]

f(x) is linear

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