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3 years ago

34. Which is the rule for the linear function graphed below?

Mathematics

1 answer:

Ivahew [28]3 years ago

8 0

Answer:

We can note that this part of the graph is a linear function. This means that is has a general form: y = mx + c where m is the slop and c is the y-intercept (value of y at x=0). For the slope, we will use the points (0,2) and (3,5) to calculate it as follows: m = (y2-y1)/(x2-x1) = (5-2)/(3-0) = 1 For the y-intercept, we can note that at x=0, the value of y is 2. This means that the equation of the first part of the graph is: y = x + 2

Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/703068-which-rules-define-the-function-graphed-below.html

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Morgarella [4.7K]

Answer:

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2 years ago

5 + 4x - 2x = 8x + 11 + 12 Explain step by step PLEASE

Yuliya22 [10]

Answer:

The solution is x = -3

Step-by-step explanation: Please brainliest!

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2 years ago

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Which is equivalent to..... algebra II engenuity

katrin2010 [14]

Answer:

The correct answer is second option option

9¹/⁸ ˣ

Step-by-step explanation:

Points to remember

Identities

ᵃ√x = = x¹/ᵃ

√x = x¹/²

(xᵃ)ᵇ = xᵃᵇ

To find the correct option

It s given that,

(⁴√9)¹/² ˣ

By using the above identities we can write,

(⁴√9)¹/²ˣ = (9¹/⁴)¹/²ˣ [ since ⁴√9 = 9¹/⁴]

= 9⁽¹/⁴ * ¹/²⁾ ˣ

= 9¹/⁸ ˣ

Therefore the correct answer is second option option

9¹/⁸ ˣ

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2 years ago

The coordinates G(7,3), H(9, 0), (5, -1) form what type of polygon?

Marrrta [24]

Answer:

Is an acute triangle

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

G(7,3), H(9, 0), I(5, -1)

step 1

Find the distance GH

substitute in the formula

$d=\sqrt{(0-3)^{2}+(9-7)^{2}}$

$d=\sqrt{(-3)^{2}+(2)^{2}}$

$GH=\sqrt{13}\ units$

step 2

Find the distance IH

substitute in the formula

$d=\sqrt{(0+1)^{2}+(9-5)^{2}}$

$d=\sqrt{(1)^{2}+(4)^{2}}$

$IH=\sqrt{17}\ units$

step 3

Find the distance GI

substitute in the formula

$d=\sqrt{(-1-3)^{2}+(5-7)^{2}}$

$d=\sqrt{(-4)^{2}+(-2)^{2}}$

$GI=\sqrt{20}\ units$

step 4

Verify what type of triangle is the polygon

we know that

If applying the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$ ----> is a right triangle

$c^{2}> a^{2}+b^{2}$ ----> is an obtuse triangle

$c^{2}< a^{2}+b^{2}$ ----> is an acute triangle

where

c is the greater side

we have

$c=\sqrt{20}\ units$

$a=\sqrt{17}\ units$

$b=\sqrt{13}\ units$

substitute

$c^{2}= (\sqrt{20})^{2}=20$

$a^{2}+b^{2}=(\sqrt{17})^{2}+(\sqrt{13})^{2}=30$

therefore

$c^{2}< a^{2}+b^{2}$

Is an acute triangle

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3 years ago

Read 2 more answers

A sandbag is released from a hot air balloon that is 400ft above the ground. Using the falling object model, h = –16t2 + s, wher

Aleonysh [2.5K]

Answer:

It takes 5 seconds for the sandbag to hit the ground

Explanation:

Given:

Height at which the sand bag is released from the hot air balloon $= -400 f t$

To find:

Time taken for the sandbag to hit the ground, t =?

Solution:

The initial height =s

Initially the height will be 0

Hence $s=0$

the falling object model equation

$h=-16 t^{2}+s$

Substituting the values we get,

$h=-16 t^{2}+s$

$400=16 t^{2}$

$t^{2}=\frac{400}{16}$

$t^{2}=25$

$t=\sqrt{25}$

$t=5 \text { seconds }$

Result:

Thus it takes 5 seconds for the sandbag to hit the ground

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3 years ago