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Sergio039 [100]
3 years ago
9

John borrows $3000 to

Mathematics
1 answer:
Goryan [66]3 years ago
7 0
The answer is $3,996

To get this you will need to multiply the monthly payments ($333) by the total number of payments (12)

$333x12= $3996

PS: You also payed $996 in intrest

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Find the circumference of the object. Use 3.14 or
shutvik [7]

Answer:

19.73

Step-by-step explanation:

look up circumference formula and click solve

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3 years ago
Need help pls and th x s​
Sedaia [141]

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times 5

Step-by-step explanation:

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3 years ago
Solve #3 Find the length of the missing side.a) A right triangle has a side length 2.4 cm and 6.5 cm. What is the length of the
Snezhnost [94]

Answer:

A) ≈ 6.93

B) ≈ 0.02

Step-by-step explanation:

A)

Formula : a² + b² = c²

2.4² + 6.5² = c²

5.76 + 42.25 = c²

c = √48.01

c ≈ 6.93

B)

Formula = c² - a² = b²

1/3² - 1/4² = b²

1/9 - 1/16 = b²

b = √7/144

b ≈ 0.02

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

3 0
3 years ago
Read 2 more answers
Consider vectors u and v, where u = (1, -1) and v = (1, 1). What is the measure of the angle between the vectors?
7nadin3 [17]

Answer:

I believe the answer is C. 90 degrees

Step-by-step explanation:

First you find the lengths of the two vectors, which are both \sqrt{2}. Then you find the angle given the cos using:   cos(∅)=\frac{u*v}{|u|*|v|} = \frac{0}{\sqrt{2}*\sqrt{2}  } = 0. Then using ∅=acos(0)=\frac{\pi }{2} = 90°

6 0
3 years ago
Read 2 more answers
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
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