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sertanlavr [38]
3 years ago
14

Answer A and B. I will give a brainliest to the best answer.

Mathematics
1 answer:
maksim [4K]3 years ago
7 0

Answer:

a) 70 degrees

b) 110 degrees

Step-by-step explanation:

Freezing point of water is 32 degrees.

Boiling point of water is 212 degrees.

Hope this helped :)

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frosja888 [35]

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Step-by-step explanation:

Its hard  to tell the width and length from this way, but here i got 104 square inches.

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A multiple regression model has ___
Salsk061 [2.6K]

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C. More than one independent variable

Step-by-step explanation:

The difference between linear regression model and multiple regression model is that the linear regression has just one independent variable which is use to determine the dependent variable. While the multiple regression model has at least two or more independent variable which is use to determine the dependent variable.

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4 years ago
On Saturday Keisha ran 2.758 kilometers. On Sunday she ran 1.78 kilometers. How Much farther did she run on Saturday than on Sun
Sonja [21]
0.978 is the answer plz tell me if it helped xD
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Pleaseeeeeeeeess help meeee ​
stich3 [128]

Answer:

7 and 1/8 is the correct answer

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6 0
3 years ago
Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt
polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt

=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k

The first integral is trivial since (\tan t)'=\sec^2t.

The second can be done by substituting u=t^2-1:

u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C

The third can be found by integrating by parts:

u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t

\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8

\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C

8 0
4 years ago
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