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IceJOKER [234]
3 years ago
14

How you get the answer of the question?

Mathematics
2 answers:
kozerog [31]3 years ago
6 0
Least common multiple
BabaBlast [244]3 years ago
4 0
The answer is A, because it’s the only number that fits into 152 of the given choices
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HELP! Can someone please explain to me how to solve this question?
Sunny_sXe [5.5K]

Answer:

(-3.1, -32.5)

Step-by-step explanation:

Set the equations equal to each other: 25x+45=15x+14

Solve for x: 10x = -31

Solve for x: x=-3.1

Plug the x value back into one of the equations to find the value of y:

y = 15 (-3.1) +14

y = -32.5

7 0
3 years ago
Read 2 more answers
Use the figure to answer the question that follows: PLEASE HELP (20 POINTS)
erma4kov [3.2K]

Answer:

hmmmmmmmmmmmmmmmmmmm let me think about it.

4 0
3 years ago
two cyclists leave towns 74 miles apart at the same time and travel towards each other. one cyclist travels 3 mi/h faster than t
gizmo_the_mogwai [7]

The rates of each cyclist are 17mi/h and 20mi/h

What is rate?

Rate is a ratio that compares two different quantities which have different units.

In this comparison, one unit is taken to be a standard unit in which the other is compared with.

Analysis:

Let the speeds of both cyclists be x,  x+3

distance travelled by x cyclist = 2x

distance travelled by (x+3) cyclist = 2(x+3)

Total distance = 2x + 2(x+3) = 74 miles

2x +2x +6 = 74

4x + 6 = 74

4x = 74 - 6

4x = 68

x = 68/4 = 17mi/h

The other cyclist speed is x +3 = 17 +3 = 20mi/h

Learn more about rate: brainly.com/question/119866

#SPJ1

6 0
2 years ago
Suppose studies indicate that the earth's vegetative mass, or biomass for tropical woodlands, thought to be about 35 kilograms p
Vanyuwa [196]

it would be that killograms in meters

5 0
3 years ago
14. Consider the function f(x, y) = yx^4 * e^x². Find fxyxyx (x, y).​
Alex787 [66]

The second partial derivatives of f exist everywhere in its domain. By Schwarz's theorem, the mixed second-order partials derivatives are equal:

f_{xy} = f_{yx}

Then

f(x,y) = y x^4 e^{x^2} \\\\ \implies f_y = x^4 e^{x^2} \\\\ \implies f_{yx} = f_{xy} = g(x) \\\\ \implies f_{xyx} = g'(x) \\\\ \implies f_{xyxy} = 0 \\\\ \implies f_{xyxyx} = \boxed{0}

We don't actually need to compute g and g' to know that they are both free of y. So when we differentiate for a second time with respect to y, the whole thing cancels out.

5 0
2 years ago
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