All categories

2 years ago

Explain why the vertical line does not represent a linear function.

Mathematics

1 answer:

Rashid [163]2 years ago

5 0

Answer:

Its my first answer but may i have a brainlist please im new

Step-by-step explanation:

You might be interested in

I need help with 1,2,3,6

laila [671]

1. 84
2. 3.36
3. 7
6. 279

3 0

3 years ago

What is the answer for the four at the bottom?

Sladkaya [172]

What am I suppose to do so I can help you?

7 0

2 years ago

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

What is the equation, in standard form, of a parabola that contains the following points? (-2,18),(0,4),(4,24)

nydimaria [60]

Y=ax² + bx + c and A(-2,18), B(0.4), C(4,24) given

1)Let's determine 1st the value of c in using the pair of B
4= a(0)² + b(0) + c , c=4 and y=ax²+bx+4

2) Use the pair of A: 18 = 4a -2b + 4===> 4a-2b=14

3) Use C pair: 24 = 16a+4b+4 ==> 16a+4b=20

4) solve the system equation in 2) and 3) and you'll find a=2 and b=-3

Hence the equation is y =2x²-3x+4

6 0

3 years ago

What is -3x^2-4x-4=0

ss7ja [257]

Answer:

<h2>no real solution</h2>

Step-by-step explanation:

$-3x^2-4x-4=0\qquad\text{change the signs}\\\\3x^2+4x+4=0\\\\\text{use the quadratic formula:}\\\\\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then an equation has two solutions:}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\text{We have}\ a=3,\ b=4,\ c=4.\\\\\text{Substitute:}\\\\b^2-4ac=4^2-4(3)(4)=16-48=-32$

8 0

3 years ago