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pashok25 [27]
3 years ago
11

Which of these skills do you think you could teach someone else? Select all that apply.

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
8 0

Answer:

There are no correct answers. Just choose what you think you can teach.

Step-by-step explanation:

You might be interested in
90 students take a test.
Anika [276]

Answer:

Step-by-step explanation:

If there are 90 students and the ratio of boys to girls is 3:2, that means 3x+2x=90

Combine like terms: 5x=90

x=18

2x (girls)=36

3x (boys)=54

1/3 of the boys scored full marks, so 54/3=18 boys scored full marks

The number of girls who scored full marks is half of the number who did not score full marks means that 1/3 of the girls scored full marks. 36/3=12, so 12 girls scored full marks.

(I don't know what the question is sorry if it's incorrect but there's not enough information)

Hope this helps!

5 0
2 years ago
If i have a pair of shoes how many shoes do i have?
crimeas [40]

This is a simple answer 2

8 0
3 years ago
Read 2 more answers
Witch one of these is correct
Lady bird [3.3K]
42:105 is the answer
4 0
3 years ago
Read 2 more answers
What is the slope of a line that is perpendicular to the line y = x 4? the slope of the line is .
SIZIF [17.4K]

Answer:-1/4

Step-by-step explanation:

Take the negative of the reciprocal for perpendicular lines. In the case of no slope (y=number) replace y with x

7 0
2 years ago
Dy/dx= y^4 and y(2)= -1. Y(-1)=
cupoosta [38]

It looks like you're asked to find the value of y(-1) given its implicit derivative,

\dfrac{dy}{dx} = y^4

and with initial condition y(2) = -1.


The differential equation is separable:

\dfrac{dy}{y^4} = dx

Integrate both sides:

\displaystyle \int \frac{dy}{y^4} = \int dx

-\dfrac1{3y^3} = x + C

Solve for y :

\dfrac1{3y^3} = -x + C

3y^3 = \dfrac1{-x+C} = -\dfrac1{x + C}

y^3 = -\dfrac1{3x+C}

y = -\dfrac1{\sqrt[3]{3x+C}}

Use the initial condition to solve for C :

y(2) = -1 \implies -1 = -\dfrac1{\sqrt[3]{3\times2+C}} \implies C = -5

Then the particular solution to the differential equation is

y(x) = -\dfrac1{\sqrt[3]{3x-5}}

and so

y(-1) = -\dfrac1{\sqrt[3]{3\times(-1)-5}} = \boxed{\dfrac12}

6 0
2 years ago
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