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elena55 [62]
3 years ago
9

What point is on the graph of f(x) = 2.5^x

Mathematics
1 answer:
Annette [7]3 years ago
8 0

Answer:

need a pic

Step-by-step explanation:

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The sum of two numbers is 58. The larger number is 6 more than the smaller number. What are the numbers?
kenny6666 [7]

Answer:

Numbers are 32, and 26.

Step-by-step explanation:

58 / 2 = 29

29 + 3 = 32 (large number)

29 - 3 = 26 (small number)

8 0
3 years ago
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Please help it's about Phythagorean Theorem! <br> I attached the image
storchak [24]
Pythagorean Theorem is:

c² = a² + b²

Let’s find the values of a and b.

Looking at the image provided in the question we have 2 number already, 70 and 120.

Let’s name the side with length 120 as a, and the side with length 70 as b.

Then we should substitute these values into the Pythagorean Theorem:


c² = 120² + 70²

c² = 14400 + 4900

c² = 19300

Then we should work out the value of c by square rooting both sides:

√c = √19300

c = 138.92444

Therefore the ball would have to be hit a total of 138.92444 units.

If you have to round, that would be 139 units.
8 0
3 years ago
What is the slope of the line through (-7.-2) and (-6,7)
Xelga [282]

Answer:

gradient = 9

Step-by-step explanation:

5 0
3 years ago
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Compare and contrast subtracting integers and subtracting whole numbers
Galina-37 [17]

Subtracting whole numbers does not include subtracting negative numbers, whereas when subtracting integers you can include negative numbers.

6 0
3 years ago
How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%
blondinia [14]

The trick is to exploit the difference of squares formula,

a^2-b^2=(a-b)(a+b)

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2

Whatever you do to the denominator, you have to do to the numerator too. So

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2

Expand the numerator:

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}

So we have

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2

But √12 = √(3•4) = 2√3, so

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}

7 0
3 years ago
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