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I am Lyosha [343]
3 years ago
13

Help pls I'm rlly bad at math

Mathematics
2 answers:
makkiz [27]3 years ago
8 0
4 is the correct answer ::))
Inessa05 [86]3 years ago
3 0
The answer to the question is 4
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What is the slope of the line that passes through the points (3,8) and (8,13) write the answer in simplest form
qwelly [4]

Answer: y=5

X=5

5/5 or 1

Step-by-step explanation:

4 0
3 years ago
If I get a 100 on one assingment and a 57 on the other, what is my average of the assignments?
MakcuM [25]

Answer:

78.5

Step-by-step explanation:

Add the two scores, and then divide this sum by 2 to obtain the average score:

\frac{100 + 57}{2}  =  \frac{157}{2}  = 78.5

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3 years ago
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Solve the equation 12x + 6y = 24 for x.
Ulleksa [173]

Answer:

The answer is x=2-(y/2)

Step-by-step explanation:

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3 years ago
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Consider an object moving along a line with the given velocity v. Assume t is time measured in seconds and velocities have units
Lady_Fox [76]

Answer:

a. the motion is positive in the time intervals: [0,2)U(6,\infty)

   The motion is negative in the time interval: (2,6)

b. S=7 m

c. distance=71m

Step-by-step explanation:

a. In order to solve part a. of this problem, we must start by determining when the velocity will be positive and when it will be negative. We can do so by setting the velocity equation equal to zero and then testing it for the possible intervals:

3t^{2}-24t+36=0

so let's solve this for t:

3(t^{2}-8t+12)=0

t^{2}+8t+12=0

and now we factor it again:

(t-6)(t-2)=0

so we get the following answers:

t=6  and t=2

so now we can build our possible intervals:

[0,2)  (2,6)  (6,\infty)

and now we test each of the intervals on the given velocity equation, we do this by finding test values we can use to see how the velocity behaves in the whole interval:

[0,2) test value t=1

so:

v(1)=3(1)^{2}-24(1)+36

v(1)=15 m/s

we got a positive value so the object moves in the positive direction.

(2,6) test value t=3

so:

v(1)=3(3)^{2}-24(3)+36

v(3)=-9 m/s

we got a negative value so the object moves in the negative direction.

(6,\infty) test value t=7

so:

v(1)=3(7)^{2}-24(7)+36

v(1)=15 m/s

we got a positive value so the object moves in the positive direction.

the motion is positive in the time intervals: [0,2)U(6,\infty)

   The motion is negative in the time interval: (2,6)

b) in order to solve part b, we need to take the integral of the velocity function in the given interval, so we get:

s(t)=\int\limits^7_0 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{0}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{7}_{0}

so now we evaluate the integral:

s=7^{3}-12(7)^{2}+36(7)-(0^{3}-12(0)^{2}+36(0))

s=7 m

for part c, we need to evaluate the integral for each of the given intervals and add their magnitudes:

[0,2)

s(t)=\int\limits^2_0 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{2}_{0}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{2}_{0}

so now we evaluate the integral:

s=2^{3}-12(2)^{2}+36(2)-(0^{3}-12(0)^{2}+36(0))

s=32 m

(2,6)

s(t)=\int\limits^6_2 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{6}_{2}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{6}_{2}

so now we evaluate the integral:

s=6^{3}-12(6)^{2}+36(6)-(2^{3}-12(2)^{2}+36(2))

s=-32 m

(6,7)

s(t)=\int\limits^7_6 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{6}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{7}_{6}

so now we evaluate the integral:

s=7^{3}-12(7)^{2}+36(7)-(6^{3}-12(6)^{2}+36(6))

s=7 m

and we now add all the magnitudes:

Distance=32+32+7=71m

7 0
3 years ago
The average low temperature in a city in August is 49 degrees Fahrenheit and the average high temperatures is 71 degrees fahrenh
Darya [45]

Low August: 9 C or 9.4 C if you need specifics

Avg. August: 22 C or 21.66 if you need specifics


6 0
3 years ago
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