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tresset_1 [31]
3 years ago
12

According to this lesson, how many ice ages have occurred? A. 2 B. 3 C. 4 D. 1

Mathematics
2 answers:
earnstyle [38]3 years ago
6 0
Hello!

There have been 5 (major) ice ages. Names: (Huronian, Cryogenian, Andean-Saharan, Karoo Ice Age, and Quaternary glaciation.) How much does it mention?

Hope l helped!
kipiarov [429]3 years ago
5 0
I believe it is C. 4
I agree with Y∅U (sorry if I spelled it wrong), there have been 5 but in the book I used for this same question, it only listed 4.
So my final answer would be C. 4
P.S. I got the same answer correct on a test from K12.

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3 years ago
You and your friend each collect rocks and fossils. Your friend collects three times as many rocks and half as many fossils as y
tino4ka555 [31]

Answer:

Number of rocks and fossils collected by you is 1 and  24 respectively

Number of rocks and fossils collected by your friend is 3 and 12 respectively

Step-by-step explanation:

Let  the number of Rocks you collect be x

Let  the number of Fossils you collect be y

Then the total number pf objects you collected will be

x + y = 25

x = 25 - y------------------------------(1)

Your friend collects three times as many rocks and half as many fossils as you.

This can be written as

3x + y(\frac{1}{2}) = 15

3x + (\frac{y}{2}) = 15-------------------(2)

Substituting (1) in (2)

3(25 -y) + (\frac{y}{2}) = 15

75 - 3y + (\frac{y}{2}) = 15

Grouping the like terms we get,

75 - 15 = 3y - (\frac{y}{2})

60= \frac{6y-y}{2})

60 \times 2= 6y-y

120= 5y

y = \frac{120}{5}

y= 24

Substituting  y value in equation(1) we get

x = 25 - 24

x= 1

Friends collects 3 times rock

so collects 3x =3(1) = 3rocks

Also he collects half as many fossils

That is

\frac{y}{2} = \frac{24}{2} =12 fossils

4 0
4 years ago
The probability of X being at most A
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Answer:

1/26

Step-by-step explanation:

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3 years ago
what are the solution(s) to the quadratic equation 50 – x2 = 0? x = ±2 x = ±6 x = ±5 no real solution
Darina [25.2K]

Consider the given quadratic equation 50-x^2=0

x^2-50=0

Comparing the given quadratic equation to the general equation ax^2+bx+c=0, we get a= 1, b=0 and c= -50

So, the solution to the quadratic equation is given by:

x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

x = \frac{-0 \pm \sqrt{0^2-4(1)(-50)}}{2}

x = \frac{\pm \sqrt{200}}{2}

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x = \pm 5\sqrt2

So, the solutions to the given quadratic equation are x = \pm 5\sqrt2.

5 0
3 years ago
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