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Rzqust [24]
2 years ago
13

jaideep builds a house and sell it for $450000. (a)he pays a tax of 1.5% of the selling price of the house.show that he pays $67

50 in tax.​
Mathematics
1 answer:
ValentinkaMS [17]2 years ago
4 0

Answer:

$450000*0.015=$6750

Step-by-step explanation:

1.5% as a decimal is 0.015

You then multiply the sale amount by the decimal to get $6750

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Without using trignometery tables find value of , cos70/sin20+cos57.cosec33 - 2cos60
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Here  is some thoughts on this:

cos 57  = sin (90-57) = sin 33 

so cos 57 . cosec 33 = cos 57 / sin 33 = sin 33 / sin33 = 1

2 cos 60 = 2 * 1/2  = 1

so the last 2 parts work out to 0 

now we have to find cos 70 / sin 20

sin 20 = cos 70  so this comes to 1


so finally the answer is 1
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How to write six billion sixty- seven million four hundred four thousand thirteen and nineteen thousandths in standard form?
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Pls solve.. URGENT!!! Will give brainliest.<br>EXPLAIN IN STEPS<br><br>links will be reported
strojnjashka [21]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Length of each side of each of these three equilateral triangle is :

  • 1.5 \:  \: cm

Solution is in attachment ~

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2 years ago
A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th
Yuri [45]

Answer:

15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

y(0)=0, y'(0)=0

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: F_{r} (t) that correspond to the force of resistance on the mass by the action of the spring and F(t) that is an external force with unknown direction (that does not specify in the enounce).

For determinate F_{r} (t) we can use Hooke's Law given by the formula F_{r} (t) = k y(t) where k correspond to the elastic constant of the spring and y(t) correspond to  the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m}  = 441.45 \frac{N}{m}

Now we apply Newton's 2nd Law and obtaint that...

F_{r} (t) ± F(t) = ma(t)

F_{r} (t) = ky(t) = 441.45y(t)

F(t) = 170 cos(5t)

m = 15 kg

a(t) = \frac{d^{2}y(t)}{dt^{2} }

Finally... 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

We know from the problem that there's not initial displacement and initial velocity, so... y(0)=0 and y'(0)=0

Finally the Initial Value Problem that models the situation describe by the problem is

\left \{ 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.

6 0
2 years ago
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