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kozerog [31]
3 years ago
11

Help me pls i’m barely starting off and i’m so confused

Mathematics
1 answer:
strojnjashka [21]3 years ago
6 0

Answer:

plane UVR and plane XUV

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Calculate the premiums for the following insurance policies. (Help!!!! Active Quiz)
Sonbull [250]
Automobile insurance: 809.50
comprehensive coverage: 64

Based on the note:
subtract 64: 809.50 - 64 = 745.50
multiply by 150% : 745.50 x 1.5 = 1,118.25
add 64: 1,118.25 + 64 = 1,182.25

The new total is 1,182.25
7 0
3 years ago
Read 2 more answers
What is (f ° g)(2) and (g ° f)(2) with the set of points f: {(0,4), (1,2), (2,0), (3,2), (4,4)} and g: {(0,4), (1,3), (2,2), (3,
kumpel [21]

Answer:

\large\boxed{(f\circ g)(2)=0}\\\boxed{(g\circ f)(2)=4}

Step-by-step explanation:

f:\{(0,4), (1,2), \underline{(2,0)}, (3,2), (4,4)\}\\\\g:\{\underline{(0,4)}, (1,3), \underline{(2,2)}, (3,1), (4,0)\}\\\\(f\circ g)(x)=f\bigg(g(x)\bigg)\\\\g(2)=2\qquad/from\ (2,2)/\\f\bigg(g(2)\bigg)=f(2)=0\qquad/from\ (2,0)/\\\\(g\circ f)(x)=g\bigg(f(x)\bigg)\\\\f(2)=0\qquad/from\ (2,0)/\\g\bigg(f(2)\bigg)=g(0)=4\qquad/from\ (0,4)/

7 0
3 years ago
A right triangle's height is 3 times the length of its base. if the area of the triangle is 600, what is the height?
Anon25 [30]
  <span>a = 1/2 x b x h 

h = 3b 

so 1/2 x b x 3b = (1/2)(3b^2) = 150 

(150 x 2)/3 = b^2 

100 = b^2 

b = square root of 100 = 10 

h = 3b = 3 x 10 = 30</span>
3 0
3 years ago
P is the point on the line 2x+y-10=0 such that the length of OP, the line segment from the origin O to P, is a minimum. Find the
nirvana33 [79]
The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula.
The distance here is: <span><span>d2</span>=(x−0<span>)^2</span>+(y−0<span>)^2
</span>                                      =<span>x^2</span>+<span>y^2
</span></span>
To minimize this function d^2 subject to the constraint, <span>2x+y−10=0
</span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x 
</span>You can substitute this in for y in the distance function and take the derivative:
<span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2]
</span></span></span></span>
d′=1/2 (5x2−40x+100)^(−1/2)   (10x−40)<span>
</span>Setting the derivative to zero to find optimal x,
<span><span>d′</span>=0→10x−40=0→x=4
</span>
This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward).
For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).
 
Then y = 10 - 2(4) = 2.
 So the point, P, is (4,2).
8 0
3 years ago
Are angles 8 and 15 <br><br> congruent <br><br> supplementary <br><br> or neither
kap26 [50]
Angles 8 and 15 are supplementary.
3 0
3 years ago
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