Which is the best estimate for (8.9x10^8)/(3.3x10^4) written in scientific notation? A. 3x10^2 B. 6x10^2 C. 3x10^4 D. 6x10^4

Mathematics

2 answers:

xenn [34]3 years ago

7 0

Esitmate:
9 /3 = 3
and
8-4 = 4
so it would be 3x10^4
answer is C

son4ous [18]3 years ago

3 0

Answer:

Option C

Step-by-step explanation:

The given expression is $\frac{8.9\times 10^{8}}{3.3\times 10^{4}}$

Now we have to solve this.

$\frac{8.9\times 10^{8}}{3.3\times 10^{4}}$

$=\frac{8.9}{3.3}\times \frac{10^{8}}{10^{4}}$

= 2.70 × 10⁽⁸⁻⁴⁾

= 2.70 × 10⁴

≈ 3 × 10⁴

Therefore, Option C is the answer.

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A friend asks to borrow $250 for 18 months and agrees to pay 12% annual simple interest. How much interest will you earn? Round

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7 0

3 years ago

SOMEONE HELP ASAPPPPPPP!!! PLEASE

Rom4ik [11]

(1) See below for a diagram. Basically, the distance on the ground from the person to the building (34 ft) is adjacent to the angle of elevation (74 degrees) and the height of the building (labeled h in the diagram) is the side opposite the angle. Since we are dealing with opposite and adjacent we use the tangent of the angle and tan = opp/adj

Specifically, $tan74= \frac{h}{34}$
$h=(34)(tan74)$ <span class="_wysihtml5-temp-placeholder"></span>
$h=118.6$ feet.

Please be sure your calculator is set to degrees (not radians) when you do this problem.

(2) Here since P & Q are complimentary it means that their sum is 90 degrees. Since this is a right triangle that means that the remaining angle (R) must be the right angle. See below for a diagram.

sin = opp/hyp. As the sin Q = 9/41 this means that 9 is the length of the side opposite Q (the side PR) and 41 is the length of the hypotenuse. This makes the remaining side (QR) 40 in length.

cos = adj/hyp. If we focus on angle P the side adjacent (next to) is 9 and the hypotenuse is 41. Thus the cos of P = 9/41.

You could have also realized that if P & Q are complimentary the sin P = cos Q and the cos P = sin Q. We were not asked about tangent but it is also the case that tan P = cot Q and cot P = tan Q.