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Katarina [22]
3 years ago
13

Can anyone help!? Points are high!

Mathematics
2 answers:
jeyben [28]3 years ago
7 0
What is the question. I cant open the attachment
patriot [66]3 years ago
4 0

total was 49.10

mileage would be 49.10 = 2.10x +5.00

 total miles =

49.10 = 2.10 +5.00

44.10 = 2.10x

x = 44.10/2.10 = 21 miles

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No pic/link just type answer pls.
Aleksandr-060686 [28]
I got 4.42 hope this helps:)
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3 years ago
Which set of ordered pairs could be generated by an exponential function?
klemol [59]

Answer: Third option.

Step-by-step explanation:

By definition, Exponential functions have the following form:

y = ab^x

Where "b" is the base (b > 0 and b\neq 1), "a" is a coefficient (a\neq 0) and "x" is the exponent.

 It is importat to remember that the "Zero exponent rule" states that  any base with an exponent of 0 is equal to 1.

Then, for an input value 0 (x=0) the output value (value of "y") of the set of ordered pairs that could be generated by an exponential function must be 1 (y=1).

You can observe in the Third option shown in the image that when x=0, y=1

 Therefore, the set of ordered pairs that could be generated by an exponential function is the set shown in the Third option.

4 0
3 years ago
Read 2 more answers
Solve each equation. Check your solution.<br> 3(n - 7) = -30
uranmaximum [27]
3(n-7)=-30
3n-21= -30
3n = -9
n= -3
5 0
3 years ago
Read 2 more answers
Evaluate cos 17x/6<br><br> ill give brainlist
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This is the answer here u go have urhvhcgchchc

8 0
3 years ago
Match each set of vertices with the type of triangle they form.
Andrew [12]

Answer:  The calculations are done below.


Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

4 0
3 years ago
Read 2 more answers
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