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3 years ago

If b = 9 cm, c = 7 cm, and m∠A = 66°, what is the approximate area of ABC?

Mathematics

1 answer:

Reptile [31]3 years ago

4 0

Area of triangle ABC can be found using the formula ¹/₂ × b × c × sin (A)

Substitute b = 9, c = 7, and m∠A = 66° into the formula gives

Area = 9 × 7 × sin 66° = 57.55

ANSWER: D

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If RX=4 and XS=9, then XT=<br> And how do you get it?

konstantin123 [22]

Answer:

XT=6 units

Step-by-step explanation:

The picture of the question is the attached figure

step 1

In the right triangle RST

Applying the Pythagorean theorem

$RS^2=RT^2+TS^2$

we have

$RS=RX+XS=4+9=13\ units$ ---> by segment addition postulate

substitute

$RT^2+TS^2=169$ ----> equation A

step 2

In the right triangle RTX

Applying the Pythagorean theorem

$RT^2=RX^2+XT^2$

we have

$RX=4\ units$

substitute

$RT^2=4^2+XT^2$

$RT^2=16+XT^2$

$XT^2=RT^2-16$ ----> equation B

step 3

In the right triangle XTS

Applying the Pythagorean theorem

$TS^2=XS^2+XT^2$

we have

$XS=9\ units$

substitute

$TS^2=9^2+XT^2$

$TS^2=81+XT^2$

$XT^2=TS^2-81$ ----> equation C

step 4

equate equation B and equation C

$TS^2-81=RT^2-16$

$TS^2-RT^2=81-16$

$TS^2-RT^2=65$ ----> equation D

step 5

Solve the system

$RT^2+TS^2=169$ ----> equation A

$TS^2-RT^2=65$ ----> equation D

Solve by elimination

Adds equation A and equation D

$RT^2+TS^2=169\\TS^2-RT^2=65\\---------\\TS^2+TS^2=169+65\\2TS^2=234\\TS^2=117$

Find the value of RT^2

$RT^2+117=169\\RT^2=52$

step 6

Find the value of XT

equation C

$XT^2=117-81\\XT^2=36\\XT=6\ units$

7 0

3 years ago

What is an algebraic expression for the word phrase?<br> the difference of a number r and 3

Ostrovityanka [42]

Answer:

r - 3

Step-by-step explanation:

The word "difference" means that it's a subtraction problem

so the difference between r and 3 is r - 3

8 0

4 years ago

Calculus 2 Master needed, show steps with partial fraction decomposition <img src="https://tex.z-dn.net/?f=%5Cint%283x%5E2-26x%2

Vladimir79 [104]

Answer:

-ln|x−5| + 2 ln(x²+4) + 3 tan⁻¹(x/2) + C

Step-by-step explanation:

The fraction will be split into a sum of two other fractions.

The first fraction will have a denominator of x − 5. The numerator will the a polynomial of one less order, in this case, a constant A.

The second fraction will have a denominator of x² + 4. The numerator will be Bx + C.

$\frac{3x^{2}-26x+26}{(x-5)(x^{2}+4)}=\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}$

Combine the two fractions back into one using the common denominator.

$\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}=\frac{A(x^{2}+4)+(Bx+C)(x-5)}{(x-5)(x^{2}+4)}$

This numerator will equal the original numerator.

$A(x^{2}+4)+(Bx+C)(x-5)=3x^{2}-26x+26\\Ax^{2}+4A+Bx^{2}-5Bx+Cx-5C=3x^{2}-26x+26\\(A+B)x^{2}+(C-5B)x+(4A-5C)=3x^{2}-26x+26$

Match the coefficients.

$A+B=3\\C-5B=-26\\4A-5C=26$

Solve the system of equations.

$A=-1\\B=4\\C=-6$

So we can rewrite the integral as:

$\int {(\frac{-1}{x-5}+\frac{4x-6}{x^{2}+4}) } \, dx$

Solving:

$\int {\frac{-1}{x-5}\, dx + \int {\frac{4x}{x^{2}+4}} \, dx - \int {\frac{6}{x^{2}+4} } \, dx$

$-\int {\frac{1}{x-5}\, dx + 2\int {\frac{2x}{x^{2}+4}} \, dx - 6\int {\frac{1}{x^{2}+4} } \, dx$

$-ln|x-5| + 2ln(x^{2}+4) - 6(\frac{1}{2} tan^{-1}(\frac{x}{2} )) + C$

$-ln|x-5| + 2ln(x^{2}+4) - 3 tan^{-1}(\frac{x}{2} ) + C$

5 0

3 years ago

Pls help......

lora16 [44]

If the two equations in parenthesizes were there own equations which ones would equal 1 and 6
(x-1) or x=1
(3x-1) or x=1/3
3(x-1) or 3x-3 or x=1
Do you understand?

5 0

3 years ago

Read 2 more answers

What is 6(10b + 2) if b=5

ycow [4]

Answer:

312

Step-by-step explanation:

you do 10 times 5 and then u get 6(50 + 2) and then distribute the 6 and then you get 300 + 12 and then it's 312 is ur answer

7 0

4 years ago