All categories

4 years ago

>Please explain your answer and I'll mark the Brainliest!

Mathematics

2 answers:

prohojiy [21]4 years ago

8 0

Answer:

B=59.58

B'==120.42

Step-by-step explanation:

For the larger triangle, by sin rule, sinB/10=sin45/8.2

sinB=10*sin45/8.2

For the small triangle on the left hand side, by sin rule, sinB'/10=sin45/8.2

sinB'=10*sin45/8.2

So sinB=sinB' but as B'>90, B and B' are supplementary

B=sin^-1(10*sin45/8.2)=sin^-1(0.8623)=59.58

B'=180-B=180-59.58=120.42

8_murik_8 [283]4 years ago

4 0

Answer:

Step-by-step explanation:

apply sine rule 2 the triangle on left:

sinB'/10=sin45deg/8.2

sinB'=10*sin45deg/8.2=0.8623

B'=59.58deg or 120.42deg

as shown in the pic, B' > 90deg.

so B'=120.42deg.

the triangle on right has 2 sides at 8.2 so it is isosceles

B=adjacent < of B'

=180-120.42

=59.58deg

You might be interested in

Find the slope of the line.<br><br> The slope is ___

dimulka [17.4K]

The slope would be 2/2

3 0

3 years ago

Read 2 more answers

What is the area of a rectangle with length 8 1/2 and width 6 inches

likoan [24]

A = 1/2 (l × w)
A = 1/2 (8 1/2 × 6)
A = 1/2 (17/2 ×12/2)
A = 1/2 (204/4)
A = 1/2 (51)
A = 25 1/2 OR 25.5 in

5 0

3 years ago

Read 2 more answers

Which line plot shows the data?

Bogdan [553]

Answer:

I think its b

Step-by-step explanation:

Dont shoot the messenger that's just my best Interpretation

8 0

3 years ago

Read 2 more answers

A right triangle has legs that are 11 inches and 13 inches long.What is the area of the triangle?

Citrus2011 [14]

Just use your right angled triangle formula.

A= $\frac{ab}{2}$

Plug n chug

A= $\frac{11*13}{2}$
A=71.5 $in^{2}$

5 0

3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

3 years ago