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4 years ago

14

Write and solve a proportion to answer the question.

1 answer:

erma4kov [3.2K]4 years ago

8 0

Let x be the percentage required
25x=12
x=0.48
x=48%
Thus the ans is 48%
hope this helps

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City Park is a right triangle with a base of 40 yd

Elenna [48]

1,000 to 1,000
that is the answer

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4 years ago

The value of a car decreases linearly with time. The car was bought new for $62,000 and had a value of $41,000 after 6 years.

guapka [62]

62000-41000=21000
21000/6=3500
3500= 5.64516129% of 62000

7 0

3 years ago

The multiplication sign in algebra is a centered dot. We do not use the multiplication cross ×, because we do not want to confus

sergejj [24]

That is true what is the question?

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4 years ago

Solve each equation for 0 x 360 sqrt 2 cosx+1=0 please hurry

jolli1 [7]

Answer:

$120\º \text{ and } 240\º$

or in radians:

$\frac{2\pi }{3} \text{ and } \frac{4\pi }{3}$

Step-by-step explanation:

From the way you wrote, you want to solve the equation

$\sqrt {2 \cos(x)+1}=0$ for $0 \leq x < 360\º$ , or in radians $[0, 2\pi)$

$\sqrt {2 \cos(x)+1}=0$

<u>Square both sides</u>

$2 \cos(x) +1= 0$

$2\cos(x)=-1$

$\cos(x)=-\frac{1}{2}$

In the Unit Circle, considering one revolution (interval $[0, 2\pi)$ ),

the values where $\cos(x)=-\frac{1}{2}$ are in Quadrant II and III.

Once

$\cos(x)= \frac{1}{2} \text{ for } x = 60\º \text{ in the Quadrant I}$

The values where

$\cos(x)=-\frac{1}{2}$ are 120º and 240º.

7 0

3 years ago

A road has a 10% grade, meaning increasing 1 unit of rise to every 10 units of run.

fenix001 [56]

The grade is the ratio of rise to run, i.e. the slope aka the tangent.

$\tan \theta = \dfrac{1}{10}$

$\theta = \arctan 0.1 \approx 5.711^\circ$

Answer: (a) 6 degrees

For part b, the road is the hypotenuse c of a right triangle whose tangent of the small angle is 1/10. The height h or rise is the side opposite the small angle.

$\sin\theta = \dfrac h c$

$h = c \sin \theta$

We could just take the sine of the angle we got but let's get it from the tangent exactly.

$\cos^2 \theta + \sin ^2 \theta = 1$

Dividing by squared cosine

$1 + \tan ^2 \theta = 1/\cos^2 \theta = 1/(1- \sin^2 \theta)$

$(1- \sin^2 \theta) = 1/(1 + \tan^2 \theta)$

$\sin^2 \theta = 1 - 1/(1 + \tan^2 \theta)$

$\sin^2 \theta = 1 - 1/(1 + (1/10)^2) = 1-1/(101/100) = 1/101$

$h = c \sin \theta = 2 \sqrt{1/101} \approx 0.199$

Answer: (b) Rise of 0.199 km

6 0

4 years ago