Mathematics, 14.11.2019 07:31 guzmangisselle

Me solve, brainlist for first to answer

Me solve,brainlist for first to answer

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Answer from: sincere21

(i) the other two sides are 6 and 6 $\sqrt{2}$

(ii) the other two sides are $\frac{4}{3} and \frac{8}{3}$

Step-by-step explanation:

(i) Sine: sin(θ) = Opposite ÷ Hypotenuse

Cosine: cos(θ) = Adjacent ÷ Hypotenuse

Tangent: tan(θ) = Opposite ÷ Adjacent

Here adjacent side = 6

opposite side = d

angle = 45°

other angles are 90° and 45°

tan (45) = Opposite ÷ Adjacent

1 = d ÷ 6

∴ d = 6 × 1 = 6

so opposite side = 6

Hypotenuse ² = opposite side ² + adjacent side²

= 6² + 6²

= 36 + 36

= 72

hypotenuse = $\sqrt{72}$

= 6 $\sqrt{2}$

the other two sides are 6 and 6 $\sqrt{2}$

(ii) here adjacent side = 4√3

angle = 30°

other angles are 90° and 60°

opposite side = d

tan ( 30) = opposite ÷ adjacent

$\frac{1}{\sqrt{3}}$ = d ÷ 4√3

$\frac{1}{\sqrt{3}}$ = d × ( $\frac{\sqrt{3}}{4}$ )

3 d = 4

therefore d = $\frac{4}{3}$

therefore opposite side = $\frac{4}{3}$

Hypotenuse ² = opposite side ² + adjacent side²

=( $\frac{4}{3}$ )² +( $\frac{4}{\sqrt{3}}$ )²

= $\frac{64}{9}$

therefore hypotenuse = $\sqrt{\frac{64}{9}}$

= $\frac{8}{3}$

the other two sides are $\frac{4}{3} and \frac{8}{3}$

Answer from: Quest

1. f(x)=x²+10x+16

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(as, a> 0 the parabola is open upward), b=10. by putting the values.

-b/2a = -10/2(1) = -5

f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

so, vertex = (-5, -9)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

now find x-intercept put y=0 in the above equation. 0= x²+10x+16

x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

2. f(x)=−(x−3)(x+1)

by multiplying the factors, the general form is f(x)= -x²+2x+3.

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

so, vertex = (1, 4)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

3. f(x)= −x²+4

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=0. by putting the values.

-b/2a = -0/2(-1) = 0

f(-b/2a)= f(0)= −(0)²+4 =4

so, vertex = (0, 4)

now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

now find x-intercept put y=0 in the above equation. 0= −x²+4

−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

4. f(x)=2x²+16x+30

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(as, a> 0 the parabola is open upward), b=16. by putting the values.

-b/2a = -16/2(2) = -4

f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

so, vertex = (-4, -2)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

5. y=(x+2)²+4

the general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

if a> 0 parabola is opened upward.

if a< 0 parabola is opened downward.

compare the given equation with general form of parabola.