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3 years ago

I'm literally crying please help lol!!!!!!!!

Mathematics

1 answer:

sveta [45]3 years ago

8 0

By linear pair theorem
AEC is congruent to BED by linear pair theorum

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How many whole groups of 2 can we make with 7?

nataly862011 [7]

Answer:

We can make 3 whole groups of 2.

Hope that helps. x

Step-by-step explanation:

2, 4, 6, 8

The closest number near to 7 is 6 so we do 6 divided by 2 which equals 3.

6 0

3 years ago

Use the Venn diagram to calculate probabilities

Shkiper50 [21]

From the Venn diagram, we can gather that there are 35 total objects (6 in both A and B; 15 in A but not B; 10 in B but not A; and 4 in neither A nor B), and we have the probabilities

$\mathbb P(A\cap B)=\dfrac6{35}$

$\mathbb P(A)=\dfrac{15+6}{35}=\dfrac{21}{35}=\dfrac35$ (this is the answer)

$\mathbb P(B)=\dfrac{10+6}{35}=\dfrac{16}{35}$

By definition of conditional probability,

$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{\frac6{35}}{\frac{16}{35}}=\dfrac6{16}=\dfrac38$

$P(B\mid A)=\dfrac{P(B\cap A)}{P(A)}=\dfrac{\frac6{35}}{\frac{21}{35}}=\dfrac6{21}=\dfrac27$

8 0

4 years ago

Read 2 more answers

The table represents a linear function

Nataly [62]

Answer:

Step-by-step explanation:

there are 10 units between -6 and -16. but to get the actual answer you would have to divide by 2 to get -3,-1,1,3.

4 0

3 years ago

Casey can do 4 math problems in 13 minutes. How many can he do in 7 minutes?

mojhsa [17]

Answer:

2.15384615385

Step-by-step explanation:

8 0

3 years ago

Bernoulli differential equation... y'+xy=xy^2

snow_lady [41]

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

Let $z=y^{-1}$ , so that $z'=-y^{-2}y'$ . Then the ODE becomes linear in $z$ with

$-z'+xz=x\implies z'-xz=-x$

Find an integrating factor:

$\mu(x)=\exp\left(\displaystyle\int-x\,\mathrm dx\right)=e^{-x^2/2}$

Multiply both sides of the ODE by $\mu$ :

$e^{-x^2/2}z'-xe^{-x^2/2}z=-xe^{-x^2/2}$

The left side can be consolidated as a derivative:

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

Integrate both sides with respect to $x$ to get

$e^{-x^2/2}z=e^{x^2/2}+C$

where the right side can be computed with a simple substitution. Then

$z=1+Ce^{x^2/2}$

Back-substitute to solve for $y$ .

$y^{-1}=1+Ce^{x^2/2}\implies y=\dfrac1{1+Ce^{x^2/2}}$

3 0

3 years ago