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Anarel [89]
2 years ago
10

Add (1.3t3 + 0.4t2 – 24t) + (8 – 18t + 0.6t2) For each term in the second polynomial, enter the letter showing where that term s

hould be placed to add the polynomials vertically.
Mathematics
1 answer:
nignag [31]2 years ago
8 0

Answer:

(1.3t^3 + 0.4t^2 - 24t) + (8 - 18t + 0.6t^2) = 1.3t^3 + t^2 -42t  + 8

Step-by-step explanation:

Given

(1.3t^3 + 0.4t^2 - 24t) + (8 - 18t + 0.6t^2)

Required

Solve

We have:

(1.3t^3 + 0.4t^2 - 24t) + (8 - 18t + 0.6t^2)

Collect like terms

1.3t^3 + 0.6t^2 + 0.4t^2 - 24t- 18t  + 8

1.3t^3 + t^2 -42t  + 8

So:

(1.3t^3 + 0.4t^2 - 24t) + (8 - 18t + 0.6t^2) = 1.3t^3 + t^2 -42t  + 8

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In-s [12.5K]
Simplifying
y = -25x + 200

Reorder the terms:
y = 200 + -25x

Solving
y = 200 + -25x

Solving for variable 'y'.

Move all terms containing y to the left, all other terms to the right.

Simplifying
y = 200 + -25x
4 0
3 years ago
Stephen rode his bike 23.4865 miles On monday and 38.243 miles on Tuesday. How many miles did he ride in all
professor190 [17]

Answer:

61.7295 (61.7) miles

If I'm reading this correctly, this is just a simple addition equation. You'd add up 23.4865 and 38.243 to get 61.7295 miles, or 61.7 miles.

Hope I helped! ☺

6 0
2 years ago
A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i
garri49 [273]

Answer:

The degrees of freedom first given by:  

df=n-1=12-1=11  

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

t_{\alpha}= 1.796

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :  

Null hypothesis:\mu \leq 5  

Alternative hypothesis:\mu > 5  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

The degrees of freedom first given by:  

df=n-1=12-1=11  

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

t_{\alpha}= 1.796

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0
3 years ago
Which ordered pair is the best estimate for the
k0ka [10]
Given the pair of simultaneous equation:
y=-2/5x-2
y=5x+1
To evaluate the solution of the system of equation from the graph, we obtain the point at which the two linear equations have intersected. taking the values of x and y from the point of intersection that will be the solution of the system of linear equations. 
In our case the straight lines have intersected at the point (-0.5, -1.75). This implies that:
x=-0.5
y=-1.75
hence the answer is a] (-0.5, -1.75)
3 0
3 years ago
The angle of the elevation of the sun is 34 degrees. Find the length, l, of a shadow cast by a tree that is 53 feet tall. Round
gulaghasi [49]
\tan{34^o}= \frac{53}{x} 
\\
\\x= \frac{53}{\tan{34^o}} 
\\
\\x \approx 78.58

8 0
3 years ago
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