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Bezzdna [24]
3 years ago
11

How far is the point (8,6) from the origin​

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
7 0
The answer is 14 away
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If you bought a stock last year for a price of $67, and it has risen 12% since then, how much is the stock worth now, to the nea
lisabon 2012 [21]

Answer:$75.04

Step-by-step explanation:

last year price=$67

12% of 67

12/100 x 67

(12 x 67) ➗ 100

804 ➗ 100=8.04

Since it has risen by 12%

There new price is 67+8.04=75.04

4 0
3 years ago
Plz help me ASAP! Answer the question in the picture below. Explain all of your steps.
ruslelena [56]

Answer:

175 hope this helps


Step-by-step explanation:

Add 72 and 113

then subtract that from 360


8 0
3 years ago
P(more than z=-2.59)is?
romanna [79]
Neither p or z ISNT A NUMBER SO DO ONE WITH A NUMBER THAT QUESTIPON IS WRONG
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7 0
3 years ago
How do I find a2 and a3 for the following geometric sequence? 54, a2, a3, 128
vlada-n [284]
The formula for the nth term of a geometric sequence:
a_n=a_1 \times r^{n-1}
a₁ - the first term, r - the common ratio

54, a_2, a_3, 128 \\ \\
a_1=54 \\
a_4=128 \\ \\
a_n=a_1 \times r^{n-1} \\
a_4=a_1 \times r^3 \\
128=54 \times r^3 \\
\frac{128}{54}=r^3 \\ \frac{128 \div 2}{54 \div 2}=r^3 \\
\frac{64}{27}=r^3 \\
\sqrt[3]{\frac{64}{27}}=\sqrt[3]{r^3} \\
\frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\
r=\frac{4}{3}

a_2=a_1 \times r= 54 \times \frac{4}{3}=18 \times 4=72 \\
a_3=a_2 \times r=72 \times \frac{4}{3}=24 \times 4=96 \\ \\
\boxed{a_2=72, a_3=96}
7 0
3 years ago
Read 2 more answers
At time​ t, the position of a body moving along the​ s-axis is sequalsnegative t cubed plus 15 t squared minus 72 t m. a. Find t
larisa [96]

Answer:

Step-by-step explanation:

Given that at time t, the position of a body moving along the​ s-axis is sequalsnegative t cubed plus 15 t squared minus 72 t m

i.e. s(t) = -t^3+15t^2-72 t

Velocity is nothing but s'(t) = derivative of s

and acceleration is s"(t) = derivative of v(t)

v(t) = -3t^2+30t-72\\=-3(t^2-10t+24)\\= -3(t-6)(t-4)

a) v(t) =0 when t = 4 or 6

b) a(t) = -6t+30

a(t) =0 when t =5

c) Distance travelled by the body from 0 to 5 would be

s(5)-s(0)\\= -5^3+15(5^2)-72(5)\\= -125+375-360\\=-110

i.e. 110 miles (distance cannot be negative)

3 0
3 years ago
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