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2 years ago

If two triangles are congruent, then the six a0parts are congruent.

Mathematics

2 answers:

disa [49]2 years ago

6 0

The answer is corresponding

Hope this helps!

Rudiy272 years ago

3 0

Answer:
If two triangles are congruent, then the six corresponding parts are congruent

Explanation:
When writing congruency, the order of writing the letters is extremely important. This is because each side/angle in the first triangle would be congruent to its corresponding side/angle in the other triangle.

Example:
If triangle ABC is congruent to triangle DEF, then:
angle A = angle D
angle B = angle E
angle C = angle F
AB = DE
BC = EF
AC = DF

Hope this helps :)

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B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6) Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago

Two companies Company A and Company B offer the same starting salary of $20,000 per year. Company A gives a raise of $1000 each

Bingel [31]

A.) A = 20000 + 1000n
B = 20000(1.04)^n

b.) For A, sn = n/2(2a + (n - 1)d)
s20 = 20/2(2(20000) + 1000(20 - 1)) = 10(40000 + 19(1000)) = 10(40000 + 19000) = 10(59000) = $590,000
For B, sn = a(r^n - 1)/(r - 1)
s20 = 20000((1.04)^20 - 1)/(1.04 - 1) = 20000(2.191 - 1)/0.04 = 20000(1.191)/0.04 = 23822.46 / 0.04 = $595,561.57

c.) Using a graphing calculator, it takes 18 years for the
total amount earned a company B is greater that the total amount of earned at company A.

5 0

3 years ago

Alana spent $18.24 on 4.8 pounds of plant food.

____ [38]

Answer: $3.8 per pound of food.

Step-by-step explanation: 18.24 divided by 4.8

4 0

2 years ago

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Malia has her $ 10.00 allowence to spend at the fall carnival. She decides to order two ice cream cones for herself and her sist

Ierofanga [76]

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