Question

name: andrea n. aguirre period: 5 scientific notation: add or subtract \21\ choose problems to solve from 3 levels of difficulty and earn a total of exactly 21 points! show all of your work. keep track of your points. and challenge yourself...youve got this!
(1 point) (2 points) (3 points)
a. (3.2×10⁴)+(2.5×10⁴) g. (7×10³)+(6×10³) m. (5.8×10⁵)+(7.2×10⁴)
b. (7.9×10⁷)-(3.3×10⁷) h. (8.5×10⁶)-(7.8×10⁶) n. (8.75×10¹⁰)-(9.5×10⁹)
c. (6×10⁻⁹)+(2.95×10⁻⁹) i. (5.95×10⁻³)+(6.4×10⁻³) o. (4×10⁻³)+(2.1×10⁻⁴)
d. (9×10⁻⁵)-(4.5×10⁻⁵) j. (3.09×10⁻⁸)-(2.8×10⁻⁸) p. (5.39×10⁻⁸)-(6.75×10⁻⁹)
e. (3.45×10⁻²)+(5.98×10⁻²) k. (4.09×10¹²)+(5.987×10¹²) q. (7.6×10⁵)+(9.01×10⁷)
f. (8.7×10¹⁰)-(6.0125×10¹⁰) l. (1.1×10⁵)-(1.073×10⁵) r. (1.49×10⁻⁷)-(8.28×10⁻⁹)

Explanation:

Step1: Factor out the common power of 10

For example, for problem A: $(3.2\times 10^{4})+(2.5\times 10^{4})=(3.2 + 2.5)\times10^{4}$

Step2: Perform the arithmetic operation on the coefficients

For problem A: $3.2+2.5 = 5.7$, so the result is $5.7\times 10^{4}$

Problem A:

$(3.2\times 10^{4})+(2.5\times 10^{4})=(3.2 + 2.5)\times10^{4}=5.7\times 10^{4}$

Problem B:

$(7.9\times 10^{7})-(3.3\times 10^{7})=(7.9 - 3.3)\times10^{7}=4.6\times 10^{7}$

Problem C:

$(6\times 10^{-9})+(2.95\times 10^{-9})=(6 + 2.95)\times10^{-9}=8.95\times 10^{-9}$

Problem D:

$(9\times 10^{-5})-(4.5\times 10^{-5})=(9 - 4.5)\times10^{-5}=4.5\times 10^{-5}$

Problem E:

$(3.45\times 10^{-2})+(5.98\times 10^{-2})=(3.45 + 5.98)\times10^{-2}=9.43\times 10^{-2}$

Problem F:

$(8.7\times 10^{10})-(6.0125\times 10^{10})=(8.7-6.0125)\times 10^{10}=2.6875\times 10^{10}$

Problem G:

$(7\times 10^{3})+(6\times 10^{3})=(7 + 6)\times10^{3}=13\times 10^{3}=1.3\times 10^{4}$

Problem H:

$(8.5\times 10^{6})-(7.8\times 10^{6})=(8.5 - 7.8)\times10^{6}=0.7\times 10^{6}=7\times 10^{5}$

Problem I:

$(5.95\times 10^{-3})+(6.4\times 10^{-3})=(5.95 + 6.4)\times10^{-3}=12.35\times 10^{-3}=1.235\times 10^{-2}$

Problem J:

$(3.09\times 10^{-8})-(2.8\times 10^{-8})=(3.09 - 2.8)\times10^{-8}=0.29\times 10^{-8}=2.9\times 10^{-9}$

Problem K:

$(4.09\times 10^{12})+(5.987\times 10^{12})=(4.09+5.987)\times 10^{12}=10.077\times 10^{12}=1.0077\times 10^{13}$

Problem L:

$(1.1\times 10^{5})-(1.073\times 10^{5})=(1.1 - 1.073)\times10^{5}=0.027\times 10^{5}=2.7\times 10^{3}$

Problem M:

First, rewrite $7.2\times 10^{4}$ as $0.72\times 10^{5}$
$(5.8\times 10^{5})+(7.2\times 10^{4})=(5.8\times 10^{5})+(0.72\times 10^{5})=(5.8 + 0.72)\times10^{5}=6.52\times 10^{5}$

Problem N:

First, rewrite $9.5\times 10^{9}$ as $0.95\times 10^{10}$
$(8.75\times 10^{10})-(9.5\times 10^{9})=(8.75\times 10^{10})-(0.95\times 10^{10})=(8.75 - 0.95)\times10^{10}=7.8\times 10^{10}$

Problem O:

First, rewrite $2.1\times 10^{-4}$ as $0.21\times 10^{-3}$
$(4\times 10^{-3})+(2.1\times 10^{-4})=(4\times 10^{-3})+(0.21\times 10^{-3})=(4 + 0.21)\times10^{-3}=4.21\times 10^{-3}$

Problem P:

First, rewrite $6.75\times 10^{-9}$ as $0.675\times 10^{-8}$
$(5.39\times 10^{-8})-(6.75\times 10^{-9})=(5.39\times 10^{-8})-(0.675\times 10^{-8})=(5.39 - 0.675)\times10^{-8}=4.715\times 10^{-8}$

Problem Q:

First, rewrite $7.6\times 10^{5}$ as $0.076\times 10^{7}$
$(7.6\times 10^{5})+(9.01\times 10^{7})=(0.076\times 10^{7})+(9.01\times 10^{7})=(0.076 + 9.01)\times10^{7}=9.086\times 10^{7}$

Problem R:

First, rewrite $8.28\times 10^{-9}$ as $0.0828\times 10^{-7}$
$(1.49\times 10^{-7})-(8.28\times 10^{-9})=(1.49\times 10^{-7})-(0.0828\times 10^{-7})=(1.49 - 0.0828)\times10^{-7}=1.4072\times 10^{-7}$

Answer:

A. $5.7\times 10^{4}$
B. $4.6\times 10^{7}$
C. $8.95\times 10^{-9}$
D. $4.5\times 10^{-5}$
E. $9.43\times 10^{-2}$
F. $2.6875\times 10^{10}$
G. $1.3\times 10^{4}$
H. $7\times 10^{5}$
I. $1.235\times 10^{-2}$
J. $2.9\times 10^{-9}$
K. $1.0077\times 10^{13}$
L. $2.7\times 10^{3}$
M. $6.52\times 10^{5}$
N. $7.8\times 10^{10}$
O. $4.21\times 10^{-3}$
P. $4.715\times 10^{-8}$
Q. $9.086\times 10^{7}$
R. $1.4072\times 10^{-7}$