All categories

3 years ago

Find the length of BC

Mathematics

1 answer:

Ivan3 years ago

6 0

Answer:

13.3650978628

Step-by-step explanation:

Angle A=180-(Angle B+C)=180-117=63

Here,

b=BC, p=AC & AB=12

Using the relation of cos,

cosx=b/h

cos27=BC/15

15cos27=BC

Using a calculator,

BC=13.3650978628

You might be interested in

consider the following equations: -x-y=1 y=x+3 if the two equations are graphed, at what point do the lines representing the two

Oksana_A [137]

Answer:

(-2, 1)

Step-by-step explanation:

consider the following equations: -x-y=1 y=x+3

For the lines to intersect, the values must be the same

-x-y=1

x + y = -1

y = -x - 1

Equating the y values

-x - 1 = x+ 3

-x-x = 3 + 1

-2x = 4

x = 4/-2

x = -2

Substitute x = -2 into y = x+3

y = -2 + 3

y = 1

Hence the required coordinate is (-2, 1)

7 0

3 years ago

Look at each situation. Is it represented by the equation -3 + 3 = 0? Determine yes or no for each. If NO, explain why not.**

lakkis [162]

Answer:

D.Jason pays $3 on the $3 he owes in fines Yes or No?

Step-by-step explanation:

6 0

3 years ago

You put $200 into an account earning 6% interest compounded yearly.

VMariaS [17]

Answer:

43.35 years

Step-by-step explanation:

From the above question, we are to find Time t for compound interest

The formula is given as :

t = ln(A/P) / n[ln(1 + r/n)]

A = $2500

P = Principal = $200

R = 6%

n = Compounding frequency = 1

First, convert R as a percent to r as a decimal

r = R/100

r = 6/100

r = 0.06 per year,

Then, solve the equation for t

t = ln(A/P) / n[ln(1 + r/n)]

t = ln(2,500.00/200.00) / ( 1 × [ln(1 + 0.06/1)] )

t = ln(2,500.00/200.00) / ( 1 × [ln(1 + 0.06)] )

t = 43.346 years

Approximately = 43.35 years

7 0

3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers

What is the solution to q+(-9)=12

Whitepunk [10]

q−9=12

Step 1: Add 9 to both sides.

q−9+9=12+9

q=21

Answer:

q=21

7 0

3 years ago

Read 2 more answers