What part of 72 is 12? Without decimals.

In the figure shown, suppose m angle ABC=n and m angle ABD=2(m angle DBC). The angle bisector of angle DBC is BE. What is m angl

Yanka [14]

Answer:

n/6

Step-by-step explanation:

If angle bisector of angle DBC is BE. What is m angle EBC, then <DBE = <EBC

Since <ABC = <ABD+<DBC

n = 2<DBC+<DBC

n = 3<DBC

<DBC = n/3

Since <DBC = 2EBC

n/3 = 2<EBC

n/6 = <EBC

Hence <EBC = n/6

4 0

3 years ago

PLEASE HELP ME I NEED THIS ANSWER FOR ALGEBRA!

Marizza181 [45]

Answer:

$radius = 6.2$

Step-by-step explanation:

so you already have the formula, which is

$\sqrt[3]{ \frac{3v}{4\pi} }$

the v represents Volume.

and 3v would be 3×volume.

$\sqrt[3]{ \frac{3 \times 1000}{4 \times \pi} }$

$\sqrt[3]{ \frac{3000}{12.56637061} }$

$\sqrt[3]{238.7324147 }$

$= 6.20350491$

to the nearest tenth place.

$= 6.2$

4 0

2 years ago

Three fewer than the sum of c and d

balandron [24]

C+d-3
Hope this helps if you were looking to put it into an expression!

4 0

3 years ago

What is an expression to represent One minus the quotient of one and a number x.

Advocard [28]

Quotient is the result of a division.

1 - 1/x

8 0

3 years ago

Use the Fundamental Theorem of Calculus to find the "area under curve" of

lozanna [386]

Answer:

$\displaystyle A = 300$

General Formulas and Concepts:

Calculus

Integrals

Definite Integrals
Area under the curve
Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

f(x) = 6x + 19

Interval [12, 15]

Step 2: Find Area

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^{15}_{12} {(6x + 19)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^{15}_{12} {6x} \, dx + \int\limits^{15}_{12} {19} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 6\int\limits^{15}_{12} {x} \, dx + 19\int\limits^{15}_{12} {} \, dx$
[Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = 6(\frac{x^2}{2}) \bigg| \limits^{15}_{12} + 19(x) \bigg| \limits^{15}_{12}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle A = 6(\frac{81}{2}) + 19(3)$
Simplify: $\displaystyle A = 300$

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

Unit: Integration

Book: College Calculus 10e

7 0

3 years ago