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BabaBlast [244]

2 years ago

11

Commemorative coins come in packs of 15, while coin holders come in packs of 25. What are the least numbers of packs you shoul

d buy in order to have the same numbers of coins and coin holders?

1 answer:

Elis [28]2 years ago

5 0

Answer:

5 Commemorative coin packs

3 coin holders

Step-by-step explanation:

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A chef buys 9 cucumbers, 18 peppers, and 21 tomatoes. How much did he spend?6 for 2 dollars for cucumbers 12 for 9 dollars for p

geniusboy [140]

6 Cucumbers = $2
$2/6 = $0.33 per cucumber
12 peppers= $9
$9/12= $0.75 per pepper
6 tomatoes= $4
$4/6= $0.66 per tomatoes
9 cucumbers= $2.97
18 peppers= $13.5
21 tomatoes=$13.86
Total= (2.97+13.5+13.86)----> $30.33

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3 years ago

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madreJ [45]

1/
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V=42.453 in^3
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3 years ago

Ethan and Chloe shared £50 in the ratio 1:4

Alexxx [7]

Answer:

10:40

Step-by-step explanation:

5 0

2 years ago

Someone help me asap please

tiny-mole [99]

Its 2,9 because thats the highest it gets which is the maximum

8 0

3 years ago

Read 2 more answers

Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

4 0

3 years ago