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Oduvanchick [21]
3 years ago
5

Use division to determine whether the ratios form a proportion.

Mathematics
2 answers:
Andrew [12]3 years ago
4 0
A. Yes. They form a proportion.

Because:    1/4,  6/24.

Reduce 6/24 to lowest terms.  6 into 6 is 1 and 6 into 24 is 4.

6/24 = 1/4.  This is same as the other 1/4.
Leya [2.2K]3 years ago
3 0
A. yes. they form a proportion
1/4 and 6/24

6/24=1/4
1/4=.75
1/4=.75
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A car rental company has two rental rates. rate 1 is $64 per day pluus $.16 per mile. rate 2 is $128 per day plus $.08 per mile.
kvv77 [185]
D is days
m is miles

0.16m + 64d (this is rate 1)
0.08m + 128d ( this is rate 2)

since you plan to rent for one day, input d as 1.

0.16m + 64(1)
0.08m + 128(1)

i hope i gave you enough hints
7 0
4 years ago
Find the equation of this line.<br> y = [ ? ]x + [<br> Simplify your answer and enter as an integer.
lys-0071 [83]

Answer:

y= 2x -2

Step-by-step explanation:

........................................

4 0
3 years ago
Read 2 more answers
If sinx = p and cosx = 4, work out the following forms :<br><br><br>​
Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16} $

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

\boxed{\sin(-x)=-\sin(x)}

\boxed{\cos(-x)=\cos(x)}

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

\sin(x)=p

\cos(x)=4

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]

Using the sum and difference identities:

\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}

\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}

\sin(\pi -x) = \sin(x)

\sin(2\pi +x)=\sin(x)

\cos(2\pi-x)=\cos(x)

Therefore,

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]

\implies [p+4] \cdot [p \cdot 4]=4p^2+16p

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16} $

8 0
3 years ago
Can some one help me with this
m_a_m_a [10]

The answer is 20 because if there is four options that you can spin on and he has spinned 80 times 8÷4 is 20

5 0
3 years ago
Read 2 more answers
There are 14 red marbles and 20 blue marbles in a bag of marbles.
sergeinik [125]

Answer:

7:10

Step-by-step explanation:

14/20= 7/10

6 0
3 years ago
Read 2 more answers
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