Ask question

Ask question

All categories

2 years ago

14

Which ratio is equivalent to 4 to 5?

1 answer:

algol [13]2 years ago

8 0

Answer:

4 / 5 = 8 / 10

Step-by-step explanation:

4:5 is a ratio which can be rewritten as 4/5.

Any number that is multiplied by both the 4 and the 5 is an equivalent ratio to 4:5.

For example, if we multiply 2 by both sides, we have 8 to 10 which is equal to 4 to 5.

You might be interested in

grigory [225]

D=M/V. You just multiply both sides by D to get DV=M, then divide V by both sides.

8 0

2 years ago

3. Your friend has eight sweatshirts. Three sweatshirts are green, one is white, and

Sveta_85 [38]

Answer:

3/14

Step-by-step explanation:

P (B then B) = 4/8 x 3/7 or 1/2 x 3/7

6 0

2 years ago

I need help plssssss

Black_prince [1.1K]

Substitute the values given.
-2-3(2) then solve
Your answer should be -8

5 0

2 years ago

The reactants of two reactions are shown.

katovenus [111]

Answer:

The answer to your question is letter B.

Step-by-step explanation:

The correct answer is letter B because both Copper and Iron as reactants have an oxidation number of zero and when they have reacted their oxidation numbers would be different from those.

A redox reaction is a reaction that involves a transfer of electrons from one atom to another and the oxidation numbers of the reactants are different from the oxidation numbers from the products.

4 0

3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

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

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago