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3 years ago

9

The scale on a map is 3 inches for every 6 1/4 miles. If the distance on the map is 5 1/2 inches, how far apart are the cities i

n real life

1 answer:

vovikov84 [41]3 years ago

4 0

I think it is 34 3/8

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If a+b=5 and a-b=4 then b^2-a^2 =

Anna007 [38]

Answer:

$a + b = 5 \\ a \: = 5 - b \\ a - b = 4 \\ 5 - b - b = 4 \\ 5 - 2b = 4 \\ 5 - 4 = 2b \\ 1 = 2b \\ \frac{1}{2} = \frac{2b}{2} \\0.5 = b \\ b = 0.5 \\ a = 5 - b \\ = 5 - 0.5 \\ = 4.5 \\ {a \: }^{2} - {b}^{2} = ( {4.5}^{2}) - ( {0.5}^{2} ) \\ = 20.25 - 0.25 \\ = 20 \\ {a}^{2} - {b}^{2} = 20$

Step-by-step explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

5 0

3 years ago

May I please receive help?<br> Please?

ANEK [815]

$\huge\colorbox{red}T\colorbox{orange}h\colorbox{yellow}a\colorbox{green}n\colorbox{blue}k\colorbox{purple}s$

6 0

2 years ago

Read 2 more answers

Find the value of x.

azamat

Answer:

98 °

Step-by-step explanation:

We know that the sum of the angles in a triangle is 180. We also know that angle DAB is 122°, meaning that angle CAB is 180-122, which is 58°. We also know that angle CBA is 24 degrees. Therefore, angle x is just 180-(58+24), which is 98

3 0

2 years ago

What is 5/8 + 3/4 divided by -2 /3 - 5/6

pentagon [3]

$\text{Hey there!}$

$\frac{5}{4}+\frac{3}{4}\div-\frac{2}{3}\ -\ \frac{5}{6}$

$\frac{5}{8} +\frac{3}{4}=\frac{11}{8}$

$\frac{-2}{3}-\frac{5}{6}=\frac{-3}{2}$

$\text{Problem becomes:}\frac{11}{8}\div\frac{-3}{2}$

$\frac{11}{8}\div\frac{-3}{2}=\frac{-11}{12}$

$\boxed{\boxed{\bf{Answer:\frac{-11}{12}}}}\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{LoveYourselfFirst:)}$

5 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago