If the width of a passport picture is 3cm and its perimeter 14cm, what is the length?

Mathematics

1 answer:

aleksklad [387]1 year ago

8 0

Answer:

The answer is 4!!

Step-by-step explanation:

A Passport picture is normally in the rectangular formation meaning, if the width of the top of the passport picture is 3, then the bottom is 3 as well.

That lets you figure out, since perimeter is all sides added together, 3+3 = 6 so 6 is both widths combine.

With this 6 in mind, if the perimeter of the entire passport is 14cm, and you have 6cm as both widths combine, now you need to find the length of this object.

Now here, since 6 is both widths combined, simply here you do 14cm - 6cm which is 8.

8 is the total length, meaning both sides added. See now that is the answer if you are finding out the total length (which is ha not the question).

Since you are trying to find only 1 side of the length, since there are 2 sides, you do 8 divided by 2, and BAM you get which is actually your answer as well!

PS. I don't even know if this is right though it should be right, use a calculator if you are doubtful :D

Happy New year!

You might be interested in

444. IU

murzikaleks [220]

Answer:

That is the third one ：

456.222

because it has the most numbers 4,5,6 and point . 2,2,2

3 0

2 years ago

Matthew haa 18 sets of baseball cards each st has 12 cards about how many baseball card does matthew have

cluponka [151]

18 x 12 = 216

216 baseball cards

6 0

2 years ago

A 50-fluid ounce bottle of laundry detergent washes 32 loads of laundry. A 100-fluid ounce bottle washes 60 loads of laundry. Ar

Alex

Answer:

I think that the answer in NO

Step-by-step explanation:

7 0

2 years ago

Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )

Zina [86]

Answer:

$\purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

Step-by-step explanation:

$\cos( \alpha ) = \frac{ \sqrt{6} }{6} = \frac{1}{ \sqrt{6} } \\ \\ \therefore \: \sin( \alpha ) = \sqrt{1 - { \cos}^{2} ( \alpha ) } \\ \\ = \sqrt{1 - \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} } \\ \\ = \sqrt{1 - {\frac{1}{ {6} }}} \\ \\ = \sqrt{ {\frac{6 - 1}{ {6} }}} \\ \\ \red{\sin( \alpha ) = \sqrt{ { \frac{5}{ {6} }}} } \\ \\ \tan( \alpha ) = \frac{\sin( \alpha ) }{\cos( \alpha ) } = \sqrt{5} \\ \\ \sin( \beta ) = \frac{ \sqrt{2} }{4} \\ \\ \implies \: \cos( \beta ) = \sqrt{ \frac{7}{8} } \\ \\ \tan( \beta ) = \frac{\sin( \beta ) }{\cos( \beta ) } = \frac{1}{ \sqrt{7} } \\ \\ \tan( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha . \tan \beta} \\ \\ = \frac{ \sqrt{5} - \frac{1}{ \sqrt{7} } }{1 + \sqrt{5} . \frac{1}{ \sqrt{7} } } \\ \\ = \frac{ \sqrt{35} - 1 }{ \sqrt{7} + \sqrt{5} } \\ \\ \purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$