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Art [367]
3 years ago
15

Vincent started driving at 45 miles per hour. After he drove 10 miles his friend Bill started driving the same route at a rate o

f 45 miles per hour. If they continue to drive at the same rate will Bill ever catch up to Vincent?
Mathematics
1 answer:
Tamiku [17]3 years ago
4 0

Answer: No

Step-by-step explanation: Vincent started 10 miles ahead so Bill will never catch up at this rate

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Select all the expressions that are equivalent to (2)^n+³
eimsori [14]

Answer:

The expressions which equivalent to  (2)^{n+3} are:

4(2)^{n+1}  ⇒ B

8(2)^{n} ⇒ C

Step-by-step explanation:

Let us revise some rules of exponent

  • a^{m} × a^{m}  = a^{m+n}
  • (a^{m})^{n} = a^{m*n}

Now let us find the equivalent expressions of  (2)^{n+3}

A.

∵ 4 = 2 × 2

∴ 4 =  2^{2}

∴  (4)^{n+2} =  (2^{2})^{n+2}

- By using the second rule above multiply 2 and (n + 2)

∵ 2(n + 2) = 2n + 4

∴  (4)^{n+2} =  (2)^{2n+4}  

B.

∵ 4 = 2 × 2

∴ 4 =  2²

∴  4(2)^{n+1} = 2² ×  (2)^{n+1}

- By using the first rule rule add the exponents of 2

∵ 2 + n + 1 = n + 3

∴   4(2)^{n+1} =  (2)^{n+3}

C.

∵ 8 = 2 × 2 × 2

∴ 8 =  2³

∴  8(2)^{n} = 2³ ×  (2)^{n}

- By using the first rule rule add the exponents of 2

∵ 3 + n = n + 3

∴  8(2)^{n} =  (2)^{n+3}

D.

∵ 16 = 2 × 2 × 2 × 2

∴ 16 = 2^{4}

∴  16(2)^{n} = 2^{4}  ×  (2)^{n}

- By using the first rule rule add the exponents of 2

∵ 4 + n = n + 4

∴  16(2)^{n} =  (2)^{n+4}

E.

(2)^{2n+3} is in its simplest form

The expressions which equivalent to  (2)^{n+3} are:

4(2)^{n+1}  ⇒ B

8(2)^{n} ⇒ C

3 0
3 years ago
What is the 6th term of the geometric sequence where a1 = -4096 and a4 = 64?
Akimi4 [234]
\bf \begin{array}{llccll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_1&-4096\\
a_2&-4096r\\
a_3&-4096rr\\
a_4&-4096rrr\\
&-4096r^3\\
&64
\end{array}\implies -4096r^3=64
\\\\\\
r^3=\cfrac{64}{-4096}\implies r^3=-\cfrac{1}{64}\implies r=\sqrt[3]{-\cfrac{1}{64}}
\\\\\\
r=\cfrac{\sqrt[3]{-1}}{\sqrt[3]{64}}\implies \boxed{r=\cfrac{-1}{4}}\\\\
-------------------------------

\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
r=-\frac{1}{4}\\
a_1=-4096\\
n=6
\end{cases}
\\\\\\
a_6=-4096\left( -\frac{1}{4} \right)^{6-1}\implies a_6=-4^6\left( -\frac{1}{4} \right)^5
4 0
3 years ago
Read 2 more answers
Find the area of the parallelogram
marshall27 [118]

Answer:

The answer is B

Step-by-step explanation:

To find the area of a parallelogram you need to multiply base by height

In this problem the base is 67 square ft and the height is 52 square feet 67 times 52 = 3484

6 0
3 years ago
What is $16.50 x 7%​
Alexus [3.1K]

Answer:

1.155

Step-by-step explanation:

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4 years ago
I need help with this
PIT_PIT [208]
2x+8y
2x+8times2
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Your final answer is 18x!
7 0
3 years ago
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