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3 years ago

If triangle ABC is congruent to triangle DEF by SAS, AB=DE, B=E, what is the missing piece of information that shows SAS?

Mathematics

1 answer:

Leviafan [203]3 years ago

7 0

BC = EF

(SAS ⇒ Side-Angle-Side)

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THE AREA OF A LAKE ON THE MAP IS 12CM SQUARE CALCULATE THE ACTRUAL AREA OF THE LAKE IN KM

den301095 [7]

Answer:

if 1cm is for 1km there are 12km

Step-by-step explanation:

it can depend on the size of the map but if there is a scale which shows how many km per cm then multiply by 12

6 0

3 years ago

Find the solution of the differential equation that satisfies the given initial condition. y' tan x = 3a + y, y(π/3) = 3a, 0 &lt

Paladinen [302]

Answer:

$y(x)=4a\sqrt{3}* sin(x)-3a$

Step-by-step explanation:

We have a separable equation, first let's rewrite the equation as:

$\frac{dy(x)}{dx} =\frac{3a+y}{tan(x)}$

But:

$\frac{1}{tan(x)} =cot(x)$

So:

$\frac{dy(x)}{dx} =cot(x)*(3a+y)$

Multiplying both sides by dx and dividing both sides by 3a+y:

$\frac{dy}{3a+y} =cot(x)dx$

Integrating both sides:

$\int\ \frac{dy}{3a+y} =\int\cot(x) \, dx$

Evaluating the integrals:

$log(3a+y)=log(sin(x))+C_1$

Where C1 is an arbitrary constant.

Solving for y:

$y(x)=-3a+e^{C_1} sin(x)$

$e^{C_1} =constant$

So:

$y(x)=C_1*sin(x)-3a$

Finally, let's evaluate the initial condition in order to find C1:

$y(\frac{\pi}{3} )=3a=C_1*sin(\frac{\pi}{3})-3a\\ 3a=C_1*\frac{\sqrt{3} }{2} -3a$

Solving for C1:

$C_1=4a\sqrt{3}$

Therefore:

$y(x)=4a\sqrt{3}* sin(x)-3a$

3 0

4 years ago

The weekly ad for a local grocery store advertises a 5 pound bag of organic apples for $13.95

masya89 [10]

There is no question therefore I cannot answer

4 0

3 years ago

Please help I have no clue what to do

max2010maxim [7]

Answer: m= 3

Step-by-step explanation:

Because 4xm=12 4x3=12 so the answer will be 3

8 0

3 years ago

I want to know how to calculate the radius, Im very confused.

11111nata11111 [884]

Answer:

a 10 ft

Step-by-step explanation:

To find the height of similar shapes, write a proportion of equal ratios.

$\frac{12}{15}=\frac{8}{y}$

Cross multiply to solve for y.

12y = 15(8)

12y = 120

y=10

4 0

3 years ago