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3 years ago

Find the area of triangle ABC with the given parts. Round to the nearest whole number.

Mathematics

1 answer:

slavikrds [6]3 years ago

3 0

$\sin{26.4^o}= \frac{h}{12.3} 
\\
\\h=12.3\times \sin{26.4^o}=5.47
\\
\\A= \frac{7.7\times5.47}{2}= 21.06$

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Triangle ABCis rotated 90° clockwise around the origin to get

Elanso [62]

Answer:

The distance A’C’ is 4.47 units

Step-by-step explanation:

Before we go on, we need to get the appropriate transformation

Mathematically, we have a 90 degrees clockwise rotation yielding the following;

(x,y) to (-y,x)

A is (-4,1)

C is (-2,5)

By transforming, we have

A’( -1,-4)

C’ (-5,-2)

To get the magnitude of the line segment, we are going to use the distance formula between points

We have this as;

D = √(x2-x1)^2 + (y2-y1)^2

D = √(-5-(-1))^2 + (-2-(-4))^2

D = √(-4)^2 + (2)^2

D = √(16 + 4)

D = √20

D = 4.47 units

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 92.3 times 4 to the power of ten

steposvetlana [31]

I am pretty sure your answer will be 96783564.8, but it may need to be rounded. If I am wrong, I am sorry.

Hope this helps~!

5 0

3 years ago

Insert parenthesis in the expression: 5+4*2+6-2*2-1 so that it has a value of 19

faust18 [17]