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3 years ago

Which of the following letters drawn on a coordinate plane would represent a function?

Mathematics

2 answers:

miv72 [106K]3 years ago

6 0

D)V

Hoped this helped!

Luden [163]3 years ago

4 0

Answer:

Step-by-step explanation:

Any vertical line passing through it would only touch it once.

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Someone Help me solve these by substitution please !

PSYCHO15rus [73]

I think the answer is:

X=3

Y=-2

8 0

2 years ago

Please help I really need help

katrin2010 [14]

Answer:

Step-by-step explanation:

Use the vertical line test, you can conclude that this is not a function.

5 0

2 years ago

Read 2 more answers

2/5 divided by 3 1/10

fiasKO [112]

2/5÷31/10=2/5*10/31=4/31

8 0

3 years ago

The string of a flying kite is being held five feet off the ground. It makes a 60 degree angle with one side parallel to the gro

VashaNatasha [74]

Answer:

Maria is right

Height=265 ft

Step-by-step explanation:

In the attached diagram, the length of the kite is BC and the height of the Kite above the ground is CE.

Now, CE=CD+DE

DE=5 ft

Next, we have to determine the length of CD.

In Triangle BCD

$sin\:60^\circ=\frac{|CD|}{300} \\|CD|=300*sin 60\\|CD|\approx260\:ft$

Therefore:

CE=260+5=265 ft

From the above, we see that <u>Maria is right.</u>

Billy erroneously applied the wrong trigonometric ratio (Cosine) which made him get a value of 150ft.

5 0

3 years ago

4. Using the Method of Completing the square, find the zeroes of the following function to the nearest

Juliette [100K]

Given:

The function is:

$F(x)=2x^2+12x+5$

To find:

The zeroes of the given function by using the Method of Completing the square.

Solution:

We have,

$F(x)=2x^2+12x+5$

It can be written as:

$F(x)=2(x^2+6x)+5$

By Method of Completing the square add and subtract square of half of coefficient of x in the parenthesis.

$F(x)=2(x^2+6x+(\dfrac{6}{2})^2-(\dfrac{6}{2})^2)+5$

$F(x)=2(x^2+6x+(3)^2)-2(3)^2+5$

$F(x)=2(x+3)^2-2(9)+5$ $[\because (a+b)^2=a^2+2ab+b^2]$

$F(x)=2(x+3)^2-18+5$

$F(x)=2(x+3)^2-13$

For zeroes, F(x)=0.

$2(x+3)^2-13=0$

$2(x+3)^2=13$

$(x+3)^2=\dfrac{13}{2}$

$(x+3)^2=6.5$

Taking square root on both sides, we get

$x+3=\pm \sqrt{6.5}$

$x=\pm \sqrt{6.5}-3$

$x\approx 2.55-3$ and $x\approx -2.55-3$

$x\approx -0.45$ and $x\approx -5.55$

Therefore, the zeroes of the given function are -0.45 and -5.55.

4 0

2 years ago