All categories

3 years ago

What else would need to be congruent to show that AABC=A DEF by the

Mathematics

1 answer:

Pepsi [2]3 years ago

7 0

Answer:

D. AC ≅ DF

Step-by-step explanation:

According to the AAS Theorem, two triangles are considered congruent to each other when two angles and a mon-included side of one triangle are congruent to two corresponding angles and a corresponding non-included side of the other.

Thus, in the diagram given:

<A and <B in ∆ABC are congruent to corresponding angles <D and <E in ∆DEF.

The only condition left to be met before we can conclude that both triangles are congruent by the AAS Theorem is for a mon-included side AC to be congruent to corresponding non-included side DF.

So, AC ≅ DF is what is needed to make both triangles congruent.

You might be interested in

A cubic inch of water weighs 0.036 pound. Which of the following rational numbers is equal to the weight of a cubic inch of wate

aniked [119]

If I were you I will think the answer is B

3 0

2 years ago

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

Set <em>u</em>: $\displaystyle u = 4x^{-2}$
[<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

<u>Step 3: Integrate Pt. 2</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0

2 years ago

Will a frying pan with a circumference of 25.12 inches fit on an electric burner with a 4-inch radius?

Leokris [45]

Answer:

Yes

Step-by-step explanation:

25.12 inches of circumference is equal to the radius of 3.998 inches

4 inches > 3.998 inches

Therefore it fits

8 0

3 years ago

What is 5/6 - blank equal 1/2

mel-nik [20]

Answer:

1/3

Step-by-step explanation:

5/6-x=1/2

x=5/6-1/2

x=5/6-3/6

x=2/6

simplify

x=1/3

8 0

3 years ago

Read 2 more answers

A circle's circumference is approximately 44 cm.

Sav [38]

<h3>Answers</h3><h3>diameter = 14</h3><h3>radius = 7</h3><h3>area = 154</h3>

these are all approximate

======================================

Explanation:

C = pi*d is the formula for the circumference. We know that C = 44 approximately, so,

C = pi*d

d = C/pi

d = 44/3.14

d = 14.01 approximately

d = 14 also approximate

take half of this to get the radius

r = d/2 = 14/2 = 7

Then use this radius to find the circle's area

A = pi*r^2

A = 3.14*7^2

A = 153.86

A = 154 rounding to the nearest whole number

7 0

3 years ago