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aev [14]
2 years ago
9

Chris drives 2.3 miles to school each day. The school is located 5.6 miles from a nearby shopping plaza. How far can Chris’ hous

e be from the shopping plaza? Explain your answer.
Mathematics
1 answer:
Simora [160]2 years ago
4 0

Answer:

7.9 miles.

Step-by-step explanation:

i just took 5.6 and added it to 2.3 because he drives 2.3 miles to school and the mall is 5.6 miles from the plaza so he would have to drive the extra 2.3 miles to get there.

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Sec squared 55 - tan squared 55
sergejj [24]

<u>Answer: </u>

sec squared 55 – tan squared 55  = 1

<u>Explanation:</u>

Given, sec square 55 – tan squared 55

We know that,

\sec \Theta=\frac{\text {hypotenuse}}{\text {base}}

And,

\tan \theta=\frac{\text { perpendicular }}{\text { base }}

where Ө is the angle

Substituting the values

\left(\frac{\text {hypotenuse}}{\text {base}}\right)^{2}-\left(\frac{\text { perpendicular }}{\text {base}}\right)^{2}

Solving,

\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}

According to Pythagoras theorem,

\text { (hypotenuse) }^{2}-\text { (perpendicular) }^{2}=(\text { base })^{2}

Putting this in the equation;

squared 55 - tan squared 55 =

\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}=\frac{(\text {base})^{2}}{(\text {base}) *(\text {base})}=1

Therefore, sec squared 55 – tan squared 55 = 1

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3 years ago
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