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2 years ago

7

A person invests 6500 dollars in a bank. The bank pays 6.5% interest compounded annually. To the nearest tenth of a year, how lo

ng must the person leave the money in the bank until it reaches 22600 dollars?

1 answer:

slamgirl [31]2 years ago

7 0

The person needs to leave it for 4 mores years to reach 22600 dollars. please give Brainliest :)

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The top and bottom margins of a poster are each 15 cm and the side margins are each 10 cm. If the area of printed material on th

12345 [234]

Answer:

the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.

Step-by-step explanation:

From the given question.

Let p be the length of the of the printed material

Let q be the width of the of the printed material

Therefore pq = 2400 cm ²

q = $\dfrac{2400 \ cm^2}{p}$

To find the dimensions of the poster; we have:

the length of the poster to be p+30 and the width to be $\dfrac{2400 \ cm^2}{p} + 20$

The area of the printed material can now be: $A = (p+30)(\dfrac{2400 }{p} + 20)$

= $2400 +20 p +\dfrac{72000}{p}+600$

Let differentiate with respect to p; we have

$\dfrac{dA}{dp}= 20 - \dfrac{72000}{p^3}$

Also;

$\dfrac{d^2A}{dp^2}= \dfrac{144000}{p^3}$

For the smallest area $\dfrac{dA}{dp }=0$

$20 - \dfrac{72000}{p^2}=0$

$p^2 = \dfrac{72000}{20}$

p² = 3600

p =√3600

p = 60

Since p = 60 ; replace p = 60 in the expression q = $\dfrac{2400 \ cm^2}{p}$ to solve for q;

q = $\dfrac{2400 \ cm^2}{p}$

q = $\dfrac{2400 \ cm^2}{60}$

q = 40

Thus; the printed material has the length of 60 cm and the width of 40cm

the length of the poster = p+30 = 60 +30 = 90 cm

the width of the poster = $\dfrac{2400 \ cm^2}{p} + 20$ = $\dfrac{2400 \ cm^2}{60} + 20$ = 40 + 20 = 60

Hence; the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.

4 0

3 years ago

Plz help me with this

never [62]

A) is right
b) just subtract ljm and kjm so 145-48 = 92°
c) 9x-25=7x+7 so here you put npq=mpq+mpn and you combine terms so it's now 2x=32 then divide by 2 on both sides and x= 16
now you just plug it into 3x-5 and you get 43°

3 0

3 years ago

Establish the identity.<br> (2 cos 0-6 sin 0)² + (6 cos 0+2 sin 0)2 = 40

kogti [31]

Rewriting the left-hand side as follows,

$(2\cos\theta-6\sin \theta)^2 +(6\cos \theta+2\sin \theta)^2\\\\=4\cos^2 \theta-24\cos \theta \sin \theta+36 \sin^2 \theta+36 \cos^2 \theta+24 \cos \theta \sin \theta+4 \sin^2 \theta\\\\=40\cos^2 \theta+40 \sin^2 \theta\\\\=40(\cos^2 \theta+\sin^2 \theta)\\\\=40$

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1 year ago

What is (3 x 20) + 9 and what is (4 x 100) + 20?

madreJ [45]

Answer:

(3 x 20) + 9 = 69

4 x 100) + 20 = 420

7 0

3 years ago

Read 2 more answers

Length width height equals to 62, what is the maximum volume

joja [24]

Assuming that the given shape is a cuboid,

volume=l×w×h

v=s³ (since cuboids have same length, they have same side length)

62=s³
∛62=s
s=3.957 ≈ 4 units

Hope I helped :)

4 0

3 years ago