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3 years ago

THIS IS VERY IMPORTANT!!!!!!!!

Mathematics

1 answer:

saul85 [17]3 years ago

6 0

I’m pretty sure it’s Abraham Lincoln, not sure though

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How do you figure this out?

pantera1 [17]

$3 ^{ - 2} = \frac{1}{3 {}^{2} } = \frac{1}{9}$

6 0

3 years ago

Read 2 more answers

Can someone solve this?

aleksley [76]

Answer:

Considering the Law of Cosines:

$c^2=a^2+b^2-2ab*cosC$

$a=14mi; b=24mi; c=CA; C=91degrees$

$\cos \left(91^{\circ \:}\right)=-0.01745\dots$

$c^2=14^2+24^2-2(14)(24)*\cos \left(91^{\circ \:}\right)\\c^2=196+576-672*\cos \left(91^{\circ \:}\right)\\c^2=772-672*\cos \left(91^{\circ \:}\right)\\c^2=772-672*(-0.01745)\\c^2=772+11.7264\\c=\sqrt{772+11.7264} \\c=\sqrt{783.7264}\\c=27.99mi$

4 0

3 years ago

The first decision is a choice between

pantera1 [17]

Answer:

min-mid-max-mid-min

-cosine

Step-by-step explanation:

This is the correct answer, further proof in the file attached.

4 0

3 years ago

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A biker traveled 24 miles in 6 hours. A unit rate to describe this would be 4 miles per hour. What is another unit rate in this

padilas [110]

Answer:

(1/4) hr/mile

Step-by-step explanation:

4 mph can be rewritten using other units of measurement:

Suppose we convert 'miles' to 'feet:'

Then:

4 miles 5280 ft

---------- * ------------- = 21120 ft/hr

1 hr 1 mile

Note that we could invert 4 mph and obtain

1 hr

-------- or (1/4) hr/mile

4 mi

4 0

3 years ago

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Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

$114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)$

Applying the following property of exponentials numbers in (II):

$e^{a}.e^{b}=e^{a+b}$

Therefore $e^{-28k}$ can be written as $e^{-14k}.e^{-14k}$

$152=190-Ae^{-14k}.e^{14k}$

Replacing (I) in the previous equation:

$152=190-76e^{-14k}$

Solving for k:

Subtracting 190 both sides, dividing by -76:

$0.5=e^{-14k}$

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

$Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76$

Dividing by 0.5

A=152

7 0

3 years ago