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Delvig [45]
2 years ago
9

Which of the following has a graph that is a straight line? (4 points) Equation 4: y = 3x3 Equation 3: y = 2x2 + 4 Equation 1: y

= 2x + 7 Equation 2: y2 = x − 1
Mathematics
2 answers:
kondor19780726 [428]2 years ago
6 0

Answer:

Equation 1: y = 2x + 7  has a graph that is a straight line

Step-by-step explanation:

Equation 4: y = 3x^3

Equation 3: y = 2x^2 + 4

Equation 1: y = 2x + 7

Equation 2: y^2 = x − 1

The graph will be a straight line for linear equation

When the largest exponent is 1 then it is a linear equation

Equation 1: y = 2x + 7 , largest exponent is 1 so it is linear

Equation 2: y^2 = x − 1, highest exponent is 2 so it is not linear

Equation 3: y = 2x^2 + 4 , highest exponent is 2 so it is not linear

Equation 4: y = 3x^3 , highest exponent is 3 so it is not linear

Equation 1: y = 2x + 7  has a graph that is a straight line

Ray Of Light [21]2 years ago
3 0
The equation of the line: y = mx + b.

Therefore your answer is Equation 1: y = 2x + 7.
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Use the distributive property to write the following expressions in expanded form. 5(7h+3m)
lilavasa [31]

Answer:

35h + 15m

Step-by-step explanation:

<u>Step 1:  Distribute</u>

5(7h + 3m)

<em>35h + 15m</em>

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Answer:  35h + 15m

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3 years ago
PLEASE!! I NEED HELP
elixir [45]

Answer:

Option A, \frac{1}{7}

Step-by-step explanation:

<u>Step 1:  Determine the numbers that are multiples of 3</u>

8 → Not a multiple of 3

9 → Multiple of 3

14 → Not a multiple of 3

19 → Not a multiple of 3

25 → Not a multiple of 3

34 → Not a multiple of 3

38 → Not a multiple of 3

<u>Step 2:  Determine the probability of chosing a multiple of 3</u>

Since we only have 1 number that is a multiple of 3 out of the 7 given numbers, that means that we do the following:  

\frac{(1\ multiple\ of\ 3)}{7\ total\ numbers}

\frac{1}{7}

Answer: Option A, \frac{1}{7}

6 0
2 years ago
The list shows the numbers of employees in the nine departments at a company. 14, 23, 6, 54, 30, 26, 17, 3, 26 What is the range
Usimov [2.4K]

Answer: 51

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Which of the following are solutions to the equation tan^2x-3=0? Check all that apply.
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If cos theta= -8/17 and theta is in quadrant 3, what is cos2 theta and tan2 theta
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\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad  &#10;\begin{array}{llll}&#10;\textit{now, hypotenuse is always positive}\\&#10;\textit{since it's just the radius}&#10;\end{array}&#10;\\\\\\&#10;thus\qquad cos(\theta)=\cfrac{-8}{17}\cfrac{\leftarrow adjacent=a}{\leftarrow  hypotenuse=c}

since the hypotenuse is just the radius unit, is never negative, so the - in front of 8/17 is likely the numerator's, or the adjacent's side

now, let us use the pythagorean theorem, to find the opposite side, or "b"

\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite&#10;\end{cases}&#10;\\\\\\&#10;\pm\sqrt{17^2-(-8)^2}=b\implies \pm\sqrt{225}=b\implies \pm 15=b

so... which is it then? +15 or -15? since the root gives us both, well
angle θ, we know is on the 3rd quadrant, on the 3rd quadrant, both, the adjacent(x) and the opposite(y) sides are negative, that means,  -15 = b

so, now we know, a = -8, b = -15, and c = 17
let us plug those fellows in the double-angle identities then

\bf \textit{Double Angle Identities}&#10;\\ \quad \\&#10;sin(2\theta)=2sin(\theta)cos(\theta)&#10;\\ \quad \\&#10;cos(2\theta)=&#10;\begin{cases}&#10;cos^2(\theta)-sin^2(\theta)\\&#10;\boxed{1-2sin^2(\theta)}\\&#10;2cos^2(\theta)-1&#10;\end{cases}&#10;\\ \quad \\&#10;tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\&#10;-----------------------------\\\\&#10;cos(2\theta)=1-2sin^2(\theta)\implies cos(2\theta)=1-2\left( \cfrac{-15}{17} \right)^2&#10;\\\\\\&#10;cos(2\theta)=1-\cfrac{450}{289}\implies cos(2\theta)=-\cfrac{161}{289}




\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\left( \frac{-15}{-8} \right)}{1-\left( \frac{-15}{-8} \right)^2}&#10;\\\\\\&#10;tan(2\theta)=\cfrac{\frac{15}{4}}{1-\frac{225}{64}}\implies tan(2\theta)=\cfrac{\frac{15}{4}}{-\frac{161}{64}}&#10;\\\\\\&#10;tan(2\theta)=\cfrac{15}{4}\cdot \cfrac{-64}{161}\implies tan(2\theta)=-\cfrac{240}{161}
6 0
3 years ago
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