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Anvisha [2.4K]
4 years ago
14

Write and solve a proportion to answer the question.

Mathematics
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
4 0
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]
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7 0
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
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