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Sidana [21]
3 years ago
13

Can someone pls help me with this?! I’m stumped lol

Mathematics
1 answer:
forsale [732]3 years ago
6 0

Answer:

  • $3.50
  • 24

Step-by-step explanation:

The graph shows you the value stored on the card dropped by $14 when Gina rented 4 videos. Thus the cost of each one is ...

... $14/4 = $3.50

Gina now has $84 on the card, so can rent an additional ...

... $84/$3.50 = 24 . . . . videos

_____

Division is a way to do "repeated subtraction." That is, if we were to subtract $3.50 from $84 repeatedly, we would find we could do it 24 times.

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If f(x) = kx^3+ x^2 − kx + 2, find a number k such that the graph of f contains the point (k, -10)
Elina [12.6K]

If we write k where we see x in the equation and set the result equal to -10, we get the result.

  • f(k)=k(k)^3+k^2-k(k)+2
  • =k^4+k^2-k^2+2
  • =k^4+2=-10
  • k^4=-12
  • k_{1}=\sqrt[4]{3}(-1-i)
  • k_{2}=\sqrt[4]{3}(-1+i)
  • k_{3}=\sqrt[4]{3}(1-i)
  • k_{4}=\sqrt[4]{3}(1+i)

4 0
1 year ago
what is a more accurate method for approximating the volume of the spherical slab other than using just cylindrical slabs.
jasenka [17]

A more accurate method for approximating the volume of the spherical slab other than using just cylindrical slabs is the; use the cross sections of the actual sphere, and to not use the cylindrical slabs.

<h3>What is a Cylinder?</h3>

A Cylinder is a surface created by projecting a closed two-dimensional curve along an axis intersecting the plane of the curve.

Sphere

This is a regular three-dimensional object in which every cross-section is a circle; the figure described by the revolution of a circle about it's diameter.

Therefore, the use of cross sections of the actual sphere is a more accurate method for approximating the volume of the spherical slab.

Learn more about cylindrical slab:

brainly.com/question/9554871

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6 0
1 year ago
A boat costs $12,050 and decreases in value by 12% per year. How much will the bout be worth after 9 Years
Mrac [35]
Value of the boat after 9 years= 12050*(1-0.12)^9
=$3814(correct to nearest integer)
8 0
3 years ago
If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =
attashe74 [19]

First of all, let <span>θθ</span> be some angle in <span><span>(0,π)</span><span>(0,π)</span></span>. Then

<span><span><span>θθ</span> is acute <span>⟺⟺</span> <span><span>θ<<span>π2</span></span><span>θ<<span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ>0</span><span>cos⁡θ>0</span></span>.</span><span><span>θθ</span> is right <span>⟺⟺</span> <span><span>θ=<span>π2</span></span><span>θ=<span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ=0</span><span>cos⁡θ=0</span></span>.</span><span><span>θθ</span> is obtuse <span>⟺⟺</span> <span><span>θ><span>π2</span></span><span>θ><span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ<0</span><span>cos⁡θ<0</span></span>.</span></span>

Now, to see if (say) angle <span>AA</span> of the triangle <span><span>ABC</span><span>ABC</span></span> is acute/right/obtuse, we need to check whether <span><span>cos∠BAC</span><span>cos⁡∠BAC</span></span> is positive/zero/negative. But what is <span><span>cos∠BAC</span><span>cos⁡∠BAC</span></span>? It is the angle made by the vectors <span><span><span>AB</span><span>−→−</span></span><span><span>AB</span>→</span></span> and <span><span><span>AC</span><span>−→−</span></span><span><span>AC</span>→</span></span>. (When you are computing the angle at a particular vertex <span>vv</span>, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex <span>vv</span> as the initial point.) We will first compute these two vectors:

<span><span><span><span>AB</span><span>−→−</span></span>=(0,0,0)−(1,2,0)=(−1,−2,0)</span><span><span><span>AB</span>→</span>=(0,0,0)−(1,2,0)=(−1,−2,0)</span></span><span><span><span><span>AC</span><span>−→−</span></span>=(−2,1,0)−(1,2,0)=(−3,−1,0)</span><span><span><span>AC</span>→</span>=(−2,1,0)−(1,2,0)=(−3,−1,0)</span></span>Therefore, the angle between these vectors is given by:<span><span><span>cos∠BAC=<span><span><span><span>AB</span><span>−→−</span></span>⋅<span><span>AC</span><span>−→−</span></span></span><span>|<span><span>AB</span><span>−→−</span></span>||<span><span>AC</span><span>−→−</span></span>|</span></span>=…</span>(1)</span><span>(1)<span>cos⁡∠BAC=<span><span><span><span>AB</span>→</span>⋅<span><span>AC</span>→</span></span><span>|<span><span>AB</span>→</span>||<span><span>AC</span>→</span>|</span></span>=…</span></span></span>Can you take it from here? From the sign of this value, you should be able to decide if angle <span>AA</span> is acute/right/obtuse.

Now, do the same procedure for the remaining two angles <span>BB</span> and <span>CC</span> as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

6 0
3 years ago
What is the equation of this line
eduard
What line? try uploading an equation
4 0
3 years ago
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