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Anna007 [38]
2 years ago
12

Find the Interest .The loan will be $25,900 over 5years at 6.25%​

Mathematics
1 answer:
koban [17]2 years ago
5 0
The interest would be $8,093.75
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Find the value of the trigonometric ratio. Express your answer as a fraction in lowest terms.
Blizzard [7]

Answer:

12/13

Step-by-step explanation:

We know that cosine is the adjacent side divided by the hypotenuse in a right triangle.

The adjacent side to angle C is BC

The hypotenuse is AC because it is opposite to the right angle

So...

cosine is (BC)/(AC)

36/39

12/13

4 0
2 years ago
Copy and complete: b 7:8:9=__:12:_
kotykmax [81]

Answer:

10.5 : 12  : 13.5

Step-by-step explanation:

7:8:9

We need to get 8 to 12

12/8 = 3/2

We need to multiply each number by 3/2

7 * 3/2:8* 3/2  :9*3/2

21/2  : 12  : 27/2

10.5 : 12  : 13.5

7 0
3 years ago
Please Help! Will mark brainliest!
damaskus [11]

Answer:

440 miles

Step-by-step explanation:

miles  ÷ gallons

132 ÷ 6 = 22

22 miles per gallon

22 × 20 = 440

3 0
3 years ago
Find the slope from (0,0) to (10,10)
juin [17]

Answer:

the slope is 1

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
NO LINKS OR FILES!
Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

(b) The velocity after 3 seconds is

v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}

(c) The particle is at rest when its velocity is zero:

\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

\displaystyle \int_0^8 |v(t)|\,\mathrm dt

By definition of absolute value, we have

|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}

7 0
2 years ago
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