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3 years ago

13

I’m a bit stuck on this question...

2 answers:

lapo4ka [179]3 years ago

4 0

Answer:

It's C :)

Step-by-step explanation:

The vertical line test is drawing lines on the function

m_a_m_a [10]3 years ago

4 0

Answer:

C.

Step-by-step explanation:

GOODLUCK

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Write words to match the expression.<br><br> 3+(4x12)

IgorLugansk [536]

Add three to the product of 4 times twelve

3 0

2 years ago

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If a quadratic function has x-intercepts at (-7,0) and 10,0), where would

madam [21]

Answer:

x = 1.5

Step-by-step explanation:

Since the axis of symmetry is between the 2 x intercepts, we can find it by finding the average of the x-coordinates of the 2 points.

(-7 + 10) / 2

3/2

= 1.5

So, the axis of symmetry is at x = 1.5

8 0

2 years ago

Find the measures of EAF DAE and CAD

sdas [7]

You have to build an equation to find the answer,

angles in a straight line add upto 180°
so your equation would be,

12x+30°=180°
-30 -30
=12x=150°
÷12 ÷12
=x=12.5°

now find the measures of the angles by using 12.5 as x.

angle EAF= 2x
=2×12.5
=25° degrees

angle DAE= 4x
=4×12.5
=50°degrees

angle CAD=6x
=6×12.5
=75°degrees

check,
30+25+50+75=180°degrees

hope this helps.......

7 0

3 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

<u>case b)</u> $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

<u>case c)</u> $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

<u>case d)</u> $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

$x^{2}-2x-4=0$

5 0

2 years ago

Read 2 more answers

If 100 bullets cost $80? how much would 1 bullet cost?

Lisa [10]

Each bullet would cost 100/80 = $1.25 if 100 bullets cost $80.

3 0

2 years ago