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docker41 [41]
3 years ago
7

Felipe received a $40.00 gift card for a photo center. He used it to buy prints that cost 8 cents each. The remaining balance, B

(in dollars), on the card after buying
x prints is given by the following function. =Bx−40.000.08x

What is the remaining balance on the card if Felipe bought
40
prints?
Mathematics
1 answer:
Neko [114]3 years ago
4 0
For better representation, I think the function is written as:

B(x) = 40.00 - 0.08x

This is much more comprehensive. The variable x represents the number of prints. So, if x=40 prints, we replace this to the x in the equation to find the balance B.

B = 40 - 0.08(40)
<em>B = $36.8</em>
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Write the equation of the line with x-intercept -2 and y-intercept -1 in slope-intercept form
eduard

The x-intercept of -2 gives us an idea that point (-2,0) if found along the line. The y-intercept of -1, tells us that point (0,-1), this also tells us that b = -1.

Now that we have two points, we can solve for slope m

\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ \text{Given two points} \\ (-2,0)\rightarrow(x_1,y_1) \\ (0,-1)\rightarrow(x_2,y_2) \\  \\ \text{Substitute} \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{-1-0}{0-(-2)} \\ m=-\frac{1}{2} \end{gathered}

Now that we have both m and b. Substitute these values to the slope intercept form

\begin{gathered} \text{Slope intercept form is} \\ y=mx+b \\ \text{where} \\ m\text{ is the slope} \\ b\text{ is the y-intercept} \\  \\ \text{Substitute the values from before and we get} \\ y=-\frac{1}{2}x-1 \end{gathered}

5 0
1 year ago
Suppose that θ is an acute angle of a right triangle and that sec(θ)=52. Find cos(θ) and csc(θ).
insens350 [35]

Answer:

\cos{\theta} = \dfrac{1}{52}

\csc{\theta} = \dfrac{52}{\sqrt{2703}}

Step-by-step explanation:

To solve this question we're going to use trigonometric identities and good ol' Pythagoras theorem.

a) Firstly, sec(θ)=52. we're gonna convert this to cos(θ) using:

\sec{\theta} = \dfrac{1}{\cos{\theta}}

we can substitute the value of sec(θ) in this equation:

52 = \dfrac{1}{\cos{\theta}}

and solve for for cos(θ)

\cos{\theta} = \dfrac{1}{52}

side note: just to confirm we can find the value of θ and verify that is indeed an acute angle by \theta = \arccos{\left(\dfrac{1}{52}\right)} = 88.8^\circ

b) since right triangle is mentioned in the question. We can use:

\cos{\theta} = \dfrac{\text{adj}}{\text{hyp}}

we know the value of cos(θ)=1\52. and by comparing the two. we can say that:

  • length of the adjacent side = 1
  • length of the hypotenuse = 52

we can find the third side using the Pythagoras theorem.

(\text{hyp})^2=(\text{adj})^2+(\text{opp})^2

(52)^2=(1)^2+(\text{opp})^2

\text{opp}=\sqrt{(52)^2-1}

\text{opp}=\sqrt{2703}

  • length of the opposite side = √(2703) ≈ 51.9904

we can find the sin(θ) using this side:

\sin{\theta} = \dfrac{\text{opp}}{\text{hyp}}

\sin{\theta} = \dfrac{\sqrt{2703}}{52}}

and since \csc{\theta} = \dfrac{1}{\sin{\theta}}

\csc{\theta} = \dfrac{52}{\sqrt{2703}}

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Step-by-step explanation:

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Answer:

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substitute in: 8-5/-7-1= 3/-8

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