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Sergeeva-Olga [200]
3 years ago
9

Layla has the following data:

Mathematics
1 answer:
ratelena [41]3 years ago
6 0
The answer would be 25 because to find the range you have to subtract the highest number with the lowest number which would be 29-25=4
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How much of the circle is shaded 1/4+3/7
Snezhnost [94]
I hope this helps you



1/4+3/7


7/7.4+3.4/7.4


7/28+12/28

19/28
4 0
3 years ago
0647grams nearest tenth
balu736 [363]
1000 is the answer! hoped I helped
8 0
3 years ago
Use the properties of logarithms to expand the following expression as much as possible. Simplify any numerical expressions that
sweet [91]

Answer:

f(x,y) = \log_{4} (x-5-\sqrt{25-6\cdot y})+\log_{4} (x-5+\sqrt{25-6\cdot y})

Step-by-step explanation:

Let be f(x,y) = \log_{4}(2\cdot x^{2}-20\cdot x +12\cdot y), this expression is simplified by algebraic and trascendental means. As first step, the second order polynomial is simplified. Its roots are determined by the Quadratic Formula, that is to say:

r_{1,2} = \frac{20\pm \sqrt{(-20)^{2}-4\cdot (2)\cdot (12\cdot y)}}{2\cdot (2)}

r_{1,2} = 5\pm \sqrt{25-6\cdot y}

The polynomial in factorized form is:

(x-5-\sqrt{25-6\cdot y})\cdot (x-5+\sqrt{25-6\cdot y})

The function can be rewritten and simplified as follows:

f(x,y) = \log_{4} [(x-5-\sqrt{25-6\cdot y})\cdot (x-5+\sqrt{25-6\cdot y})]

f(x,y) = \log_{4} (x-5-\sqrt{25-6\cdot y})+\log_{4} (x-5+\sqrt{25-6\cdot y})

3 0
3 years ago
One thousand grams is <br> A) 1 milligram <br> B) 1 centigram <br> C) 1 hectogram <br> D) 1 kilogram
kirill [66]

1,000 grams = 1 kilogram

Hopefully I was helpful.

7 0
3 years ago
Find the distance, in feet, a particle travels in its first 2 seconds of travel, if it moves according to the velocity equation
just olya [345]

Answer:

The particle will travel 6 feet in first 2 seconds.

Step-by-step explanation:

We have been given that a particle moves according to the velocity equation v(t)= 6t^2-18t+12. We are asked to find the distance that the particle will travel in its first 2 seconds.

s(t)=\int |v(t)|dt

s(t)=\int\limits^2_0 |6t^2-18t+12|dt

Now, we will eliminate the absolute value sign as:

s(t)=\int\limits^1_0 6t^2-18t+12dt+\int\limits^2_1 -6t^2+18t-12dt

s(t)=[\frac{6t^3}{3}-\frac{18t^2}{2}+12t]^1_0 +[\frac{-6t^3}{3}+\frac{18t^2}{2}-12t]^2_1

s(t)=[2t^3-9t^2+12t]^1_0 +[-2t^3+9t^2-12t]^2_1

s(2)=2(1)^3-9(1)^2+12(1)-(2(0)^3-9(0)^2+12(0))-2(2)^3+9(2)^2-12(2)-(-2(1)^3+9(1)^2-12(1))

s(2)=2-9+12-(0)-16+36-24-(-2+9-12)

s(2)=5-(0)-4-(-5)

s(2)=5-4+5

s(2)=6

Therefore, the particle will travel 6 feet in first 2 seconds.

   

 

7 0
3 years ago
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