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zepelin [54]
3 years ago
9

The quotient of -18 and the sum of -2 and d

Mathematics
1 answer:
padilas [110]3 years ago
4 0
<span>sum of -2 and d: d - 2
</span><span>The quotient of -18 and the sum of -2 and d: - 18 / (d - 2)

answer

</span>  - 18
---------
  d - 2
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Which similarity statement is true for the triangles shown?<br><br> - ABC = DEF<br> - ABC = FED
kodGreya [7K]

Answer:

I’m pretty sure it’s Triangle ABC is congruent to Triangle DEF.

Step-by-step explanation:

If you rotate Triangle DEF to where the 90 degree angle is on the bottom right, you can the compare the two triangles. You can divide 300/25, 240/20, and 180/15. They all result in 12 meaning they are congruent and similar in that way. (I’m only 90% this is right btw)

3 0
3 years ago
Write an expression in simplest form for the perimeter of each figure.
Vlad [161]

Answer: P=5y+11x


Step-by-step explanation:

To solve this problem you must apply the proccedure shown below:

- By definition, the perimeter of the figure is the sum of the lenght of each side. Then, you must make the addition as following:

P=2.8y+2x+2.2y+5x+4x

Where P is the perimeter of the figure.

- Add the like terms, then, you obtain the following expression in simplest form:

P=5y+11x


8 0
4 years ago
Math question down below
tensa zangetsu [6.8K]
The mean is the average.
To solve for the average we add all the numbers up and divide by how many numbers there are.
21, 23, 25, 25, 26, 28, 28, 28, 31, 33.
268 total, now divide by how many numbers there are.
268/10
Your mean is 26.8
8 0
3 years ago
Read 2 more answers
Vx4=4 what is the answer
Volgvan

Step-by-step explanation:

v=4÷4

v=1 Ans

Do by transposition

hope it helps

8 0
3 years ago
Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.
Mama L [17]

Multiplying both sides by y^2 gives

xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1

so that substituting v=y^3 and hence \frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx} gives the linear ODE,

\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1

Now multiply both sides by 3x^2 to get

x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2

so that the left side condenses into the derivative of a product.

\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2

Integrate both sides, then solve for v, then for y:

x^3v=\displaystyle\int3x^2\,\mathrm dx

x^3v=x^3+C

v=1+\dfrac C{x^3}

y^3=1+\dfrac C{x^3}

\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}

6 0
3 years ago
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