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2 years ago

HURRY PICTURE INCLUDED

Mathematics

1 answer:

creativ13 [48]2 years ago

7 0

The highlighted part is called <u>Ray</u>.

By the way, In geometry, a ray can be defined as a part of a line that has a fixed starting point but no endpoint.

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Relate a vertical line test to understanding functions.

xenn [34]

Answer:

Each vertical line drawn through the graph will intersect a relation in only one location

Step-by-step explanation:

We know that a vertical line test is done to test whether the graph or the given relation is a function or not and it states that:

Each vertical line parallel to the co-domain drawn on a graph will intersect the graph exactly once.

i.e. there is a unique image of each pre-image.

Hence, option: C is the correct answer.

Each vertical line drawn through the graph will intersect a relation in only one location.

6 0

3 years ago

Domain and range? Typically I understand domain and range but these graphs are vague in detail. If someone could help that'd be

brilliants [131]

What do you mean? Oh oh oh I used to do this, what grade you it?

6 0

3 years ago

The Smith family is driving to a vacation spot. The number of miles they have left is 540-60x, where x represents the number of

kotegsom [21]

Answer:

Step-by-step explanation:

6 0

3 years ago

john needed 13.71 feet of wire to fix a lamp he bought 14 half feet of wire how much did he have left ver

motikmotik

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Just subtract the two values:

14.5 - 13.71 = 0.79

He had 0.79 feet of wire left

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

8 0

3 years ago

Read 2 more answers

Solve using the quadratic formula 3x^2+11x+5=0

storchak [24]

Answer:

$3x^2 + 11x + 5 = 0 : x = \frac{-11 + \sqrt{61} }{6} , x = \frac{-11 - \sqrt{61} }{6}$

Decimal:

$(x = -0.53162..., x = -3.13504...)$

Step-by-step explanation:

$3x^2+11x+5=0$

Solve with the quadratic formula:

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are:

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

For $a=3,\:b=11,\:c=5:\quad x_{1,\:2}=\frac{-11\pm \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}$

$x=\frac{-11 + \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3} : \frac{-11 + \sqrt{61} }{6}$

$x=\frac{-11 - \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3} : \frac{-11 - \sqrt{61} }{6}$

The solutions to the quadratic equation are:

$x=\frac{-11+\sqrt{61}}{6},\:x=\frac{-11-\sqrt{61}}{6}$

Hope I helped. If so, may I get brainliest and a thanks?

Thank you, have a good day! =)

8 0

3 years ago