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3 years ago

Are the triangles congruent? if they are, state how they are congruent.

Mathematics

2 answers:

Juliette [100K]3 years ago

8 0

Answer: Yes

Step-by-step explanation: by SAS

bija089 [108]3 years ago

7 0

Answer:

Step-by-step explanation:

ASA postulate of congruence.

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d1i1m1o1n [39]

What does if mean?

Step-by-step explanation:

if means not true but what if it was a question like that or a thing you ask your friends

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3 years ago

The perimeter of an equilateral triangle is 15 inches more than the perimeter of a square, and the side of the triangle is 7 inc

kykrilka [37]

The side of the triangle is 13 inches
Side of square. 6 inches

X= side of square

4x+15 =3(x+7)
4x+15=3x+21
x=6

6+7= 13

Perím square 6x4= 24
Perim triangle 13x3=39
39-24=15

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3 years ago

James is playing his favorite game at the arcade. After playing the game 3 times, he has 8 tokens remaining. He initially had 20

lapo4ka [179]

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3 years ago

If w = 5 cos (xy) − sin (xz) and x = 1/t , y = t, z = t^3 ; then find dw/dt

Scrat [10]

In this question, we find the derivatives, using the chain's rule.

Doing this, the derivative is:

$\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}$

Chain Rule:

Suppose we have a function $w(x,y,z), x = x(t), y = y(t), z = z(t)$ , and want to find it's derivative as function of t. It will be given by:

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

Thus, we have to find the desired derivatives, which are:

w of x:

$\frac{dw}{dx} = -5y\sin{(xy)} - z\cos{(xz)}$

Considering $x = \frac{1}{t}, y = t, z = t^3$

$\frac{dw}{dx} = -5t\sin{(1)} - t^3\cos{(t^2)}$

w of y:

$\frac{dw}{dy} = -5x\cos{(xy)}$

Considering $x = \frac{1}{t}, y = t$

$\frac{dw}{dy} = -\frac{5}{t}\cos{1}$

w of z:

$\frac{dw}{dz} = -x\cos{(xz)}$

Considering $x = \frac{1}{t}, z = t^3$

$\frac{dw}{dz} = -\frac{1}{t}\cos{(t^2)}$

Derivatives of x, y and z as functions of t:

$\frac{dx}{dt} = -\frac{1}{t^2}$

$\frac{dy}{dt} = 1$

$\frac{dz}{dt} = 3t^2$

Derivative of w as function of t.

Now, we just replace what we found into the formula. So

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

$\frac{dw}{dt} = (-5t\sin{(1)} - t^3\cos{(t^2)})(-\frac{1}{t^2}) - (\frac{5}{t}\cos{1}) - (\frac{1}{t}\cos{(t^2)})3t^2$

Applying the multiplications:

$\frac{dw}{dt} = \frac{5}{t}\sin{1} + t\cos{t^2} - \frac{5}{t}\cos{1} - 3t\cos{t^2}$

Applying the simplifications:

$\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}$

Which is the derivative.

For more on the chain rule, you can check brainly.com/question/12795383

8 0

3 years ago

Can someone please help me???? I dont understand AT ALL

Jlenok [28]

Answer:

<h2>In the attachment.</h2>

6 0

3 years ago