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2 years ago

Help two helpless triangles?

Mathematics

2 answers:

mezya [45]2 years ago

5 0

Answer:

1) they are similar by AAA postulate

2) 9.6 cm

Explanation:

Triangles CAB and CED are similar

because:

1) both have the same angle C

(vertically opposite angles)

2) Angle A = Angle E

(alternate angles)

3) Angle B = Angle D

(alternate angles)

CA/CE = AB/ED

10/8 = 12/DE

DE = 12 × 8/10

DE = 9.6 cm

Illusion [34]2 years ago

3 0

Answer:

see below

Step-by-step explanation:

Angle ACE = DCE since they are vertical angles

Angle A = Angle E parallel lines cut by a transversal form congruent alternate interior angles

Angle B = Angle D parallel lines cut by a transversal form congruent alternate interior angles

When all three angles are equal, the triangles are similar.

We can use ratios to find DE

DE CE

------- = ----------

AB AC

DE 8

------- = ----------

12 10

Using cross products

10 DE = 8*12

10 DE = 96

Divide by 10

DE = 9.6

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Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$