Ask question

Ask question

All categories

miss Akunina [59]

3 years ago

13

Given the sides of the four triangles below, which is NOT a scalene triangle?

2 answers:

posledela3 years ago

6 0

Answer:

under the circustances of this trajectory the triangular polverinization oxiding cells are answer choice A thus that answer is not a scalene triangle

Step-by-step explanation:

yaroslaw [1]3 years ago

6 0

Answer:

The answer to this question is C.

Step-by-step explanation:

This is because a scalene triangle MUST always have all 3 sides with different angles. If two of them are equal, then it is an icoscelese triangle. And if all 3 sides are equal then it is an equilatiral triangle.

You might be interested in

What are each step in the process for solving an equation

Damm [24]

Answer:

1. Simplify each side of the equation by removing parentheses and combining like terms.

2. Use addition or subtraction to isolate the variable term on one side of the equation.

3. Use multiplication or division to solve for the variable.

3 0

4 years ago

5/10 in the lowest terms

kramer

Answer:

1/2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Pls help giving brainliest Question: a ÷ b when a = 10 and b = 4 a ÷ b = hint its not 2.5

Ulleksa [173]

The answer is 2.5, but if you say it isn't then maybe it can be as a fraction $\frac{10}{4}$ or $\frac{5}{2}$ .

7 0

2 years ago

One Family spent $2,000 to start an online business selling goods. They now make $1,240,000 each year. If they made a one-time i

RoseWind [281]

1,240,00 times 5 years equals 6,200,000

8 0

4 years ago

Read 2 more answers

A Bernoulli differential equation is one of the form dxdy+P(x)y=Q(x)yn Observe that, if n=0 or 1, the Bernoulli equation is line

aleksandr82 [10.1K]

It's unclear whether the ODE is

$xy'+y=4xy^2$

or

$xy+y'=4xy^2$

If the first case, then divide through both sides by $y^2$ to get

$xy^{-2}y'+y^{-1}=4x$

then substitute $z=y^{-1}$ , so that $z'=-y^{-2}y'$ . Then

$-xz'+z=4x\implies-\dfrac1xz'+\dfrac1{x^2}z=\left(-\dfrac1xz\right)'=\dfrac4x$

$\implies-\dfrac1xz=\displaystyle\int\frac4x\,\mathrm dx$

$\implies z=-4x\ln|x|+Cx$

$\implies\boxed{y=\dfrac1{Cx-4x\ln|x|}}$

If the second, then dividing through by $y^2$ gives

$xy^{-1}+y^{-2}y'=4x$

The same substitution as before gives

$xz-z'=4x\implies e^{-x^2/2}z'-xe^{-x^2/2}z=\left(e^{-x^2/2}z\right)'=-4xe^{-x^2/2}$

$\implies e^{-x^2/2}z=\displaystyle-4\int xe^{-x^2/2}\,\mathrm dx$

$\implies z=4+Ce^{x^2/2}$

$\implies\boxed{y=\dfrac1{4+Ce^{x^2/2}}}$

5 0

3 years ago