All categories

3 years ago

Given that line AD is the angle bisector of

Mathematics

1 answer:

Julli [10]3 years ago

7 0

Step-by-step explanation:

3x+1 = 5x-1

3x+2= 5x

2x= 2

x= 1

You might be interested in

The area of a rectangle is greater than or equal to 154 square centimeters. The width of the rectangle is 12 centimeters and the

soldier1979 [14.2K]

Answer:

x ≤ 12.83

(Any number under or equal to 12.8333...)

Step-by-step explanation:

154 ≥ 12x

12.833.... ≥ x

x ≤ 12.83

5 0

2 years ago

Amir stands on a balcony and throws a ball to his dog, who is at ground level. The ball's height (in meters above the ground), x

kumpel [21]

Answer:

12m

Step-by-step explanation

If the height of the ball after x seconds be modelled by the equation

h(x)=−(x−2)² +16

The height of the ball at the time it is thrown will be the height at the initial time. At that point that it is initially thrown the time is 0seconds i.e x = 0

To get the height at t x = 0seconds, we will substitute x = 0 into the modeled function to have;

h(0) = -(-0-2)²+16.

h(0) = -(-2)²+16

h(0) = -4+16

h(0) = 12

The height of the ball at the time the ball is thrown is 12m

4 0

3 years ago

Read 2 more answers

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$