Simplify the expression -4(3a-1)

Pls answer this question with steps (proof).

Tems11 [23]

Answer:

Step-by-step explanation:

Since triangle BCE is a right angle triangle, we would determine angle BEC by applying the tangent trigonometric ratio. Therefore,

Tan BEC = 6/3 = 2

Angle BEC = Tan^-1(2)

Angle BEC = 63.4°

The sum of the angles on a straight line is 180°. This means that

Angle AED + angle DEC + angle BEC = 180

Angle AED = 180 - (45 + 63.4) = 71.6°

Angle ADE = angle AED = 71.6°

Angle CDE + angle ADE = 180(sum of angles on a straight line)

Angle CDE = 180 - 71.6 = 108.4°

To get line EC, we would apply Pythagoras theorem. Therefore

EC² = 3² + 6² = 45

EC = √45 = 6.71 cm

The sum of the angles in a triangle is 180°

Therefore,

Angle ECD = 180 - (45 + 108.4) = 26.6°

By applying sine rule,

6.71/sin108.4 = ED/sin26.6 = DC/Sin45

6.71/sin108.4 = ED/sin26.6

Cross multiplying, it becomes

6.71sin26.6 = EDsin108.4

ED = 6.71sin26.6/sin108.4

ED = 3.00608/0.949 = 3.18cm

The area of a triangle is

Area = 1/2abSinC

Therefore, area of triangle EDC = 1/2 ×

ED × EC × SinDEC

Area = 1/2 × 6.71 × 3.18 × sin45

Area = 1/2 × 6.71 × 3.18 × 0.707

Area = 7.54 cm²

6 0

3 years ago

Find the equation of the parabola y = ax2 + bx + c that passes through the points (0, 3), (1,4), and (2, 3).

Temka [501]

D is the correct answer

6 0

4 years ago

<<br> Nearpod: Make X<br> c<br> 1) 2x – 5 = 19

Neko [114]

Answer:

x=12

Hope This Helps

6 0

3 years ago

Solve this equation 3/8n=-1/4

lianna [129]

(3/8)*n = -1/4 // + -1/4

(3/8)*n-(-1/4) = 0

(3/8)*n+1/4 = 0

3/8*n+1/4 = 0 // - 1/4

3/8*n = -1/4 // : 3/8

n = -1/4/3/8

n = -2/3

n = -2/3

6 0

3 years ago

An equation of a hyperbola is given.

siniylev [52]

Answer:

a)

The vertices are $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

The foci are $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$ .

The asymptotes are $y=2x,\:y=-2x$ .

b) The length of the transverse axis is 6.

c) See below.

Step-by-step explanation:

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$ is the standard equation for a right-left facing hyperbola with center $\left(h,\:k\right)$ .

a)

The vertices $\:\left(h+a,\:k\right),\:\left(h-a,\:k\right)$ are the two bending points of the hyperbola with center $\:\left(h,\:k\right)$ and semi-axis a, b.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and vertices $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

For a right-left facing hyperbola, the Foci (focus points) are defined as $\left(h+c,\:k\right),\:\left(h-c,\:k\right)$ where $c=\sqrt{a^2+b^2}$ is the distance from the center $\left(h,\:k\right)$ to a focus.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ $c=\sqrt{3^2+6^2}= 3\sqrt{5}$ and foci $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$

The asymptotes are the lines the hyperbola tends to at $\pm \infty$ . For right-left hyperbola the asymptotes are: $y=\pm \frac{b}{a}\left(x-h\right)+k$

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and asymptotes

$y=\frac{6}{3}\left(x-0\right)+0,\:\quad \:y=-\frac{6}{3}\left(x-0\right)+0\\y=2x,\:\quad \:y=-2x$

b) The length of the transverse axis is given by $2a$ . Therefore, the lenght is 6.

c) See below.

4 0

4 years ago