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3 years ago

A triangle ABC has two side lengths of 28 ft. and 24 ft. If angle C is 91°, what's the measurement of angle B

Mathematics

1 answer:

Zina [86]3 years ago

3 0

Answer:

Step-by-step explanation:

Given that in a triangle ABC, AB = 28, BC = 24 and angle C =91,

We have to find measure of angle B.

Since three measurements are given, we can solve the triangle easily

Use the sine formula for triangles.

$\frac{c}{SinC} =\frac{b}{Sin B}=\frac{a}{Sin A}$

$A = arcsin(\frac{asinC}{c} )=58.98$

Since sum of 3 angles is 180, we calculate B easily

Angle B = 180-(A+C)

= 180-(91+58.98)

=30.02

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maria [59]

Answer:

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Step-by-step explanation:

Here is the correct answer

Heather puts together 2/5 of a puzzle on Saturday and 3/10 on Sunday. Nicole has the same puzzle, and she puts together 4/5 of her puzzle.How much more of the puzzle does Nicole put together than Heather? A. 1/10 b. 3/10 c. 1/5 d 4/5

total puzzle heather put together =

$\frac{2}{5} + \frac{3}{10}$

$\frac{4 + 3}{10}$ = 7/10

the next step is to subtract 7/10 from the fraction Nicole put together

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AleksAgata [21]

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3 years ago

Read 2 more answers

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$