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3 years ago

Juan will put fencing around the outside of the enclosure. How much fencing does he need for the enclosure?

Mathematics

1 answer:

adell [148]3 years ago

3 0

If Juan intends to put fencing around the enclosure, that means that we are looking at the perimeter of the image (the outside). We can count the amount of lines between the dots to determine how much fencing he will need. After counting the perimeter of this figure, we see that the total is 22. We do not know what the units are, so for now we can just say that he will need 22 units of fencing to fully enclose this shape.

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A train to New York city leaves a station every 7 minutes. Another train to Boston leaves the station every 6 minutes. Suppose i

Virty [35]

Answer:

7:12 a.m

Step-by-step explanation:

Given that:

Train to New York city leaves station every 7 minutes

Train to Boston leaves station every 6 minutes

Current time = 6:30 a.m

Time both will leave station together is obtained by finding the least common multiple of 6 and 7

Multiples of ;

6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60,...

7 = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70,...

The least common multiple of 6 and 7 is 42.

Hence, both trains will leave the station together after 42 minutes ;

Current time + 42 minutes

6:30 a.m+ 42 minutes = 7:12 a.m

7 0

3 years ago

Find vertical and horizontal asymtope for (x^2+2x-48)/x+8

bonufazy [111]

There are none.

$y=\dfrac{x^{2}+2x-48}{x+8}=\dfrac{(x+8)(x-6)}{x+8}\\\\y=x-6$

The expression describes a straight line with a hole at x=-8 where it is undefined.

5 0

3 years ago

10 points

aivan3 [116]

Answer:

20.18

Step-by-step explanation:

Greatest number= 35.18

Least number=15

Difference between the twain=20.18

5 0

3 years ago

Read 2 more answers

let a = (a1, a2) and b = (b1, b2) and c = (c1,c2) be three non zero vectors. if a1b2 - a2b1 is not equal to 0. then show three a

Ksenya-84 [330]

Consider the contrapositive of the statement you want to prove.

The contrapositive of the logical statement

p ⇒ q

¬q ⇒ ¬p

In this case, the contrapositive claims that

"If there are no scalars α and β such that c = αa + βb, then a₁b₂ - a₂b₁ = 0."

The first equation is captured by a system of linear equations,

$\begin{cases}c_1 = \alpha a_1 + \beta b_1\\ c_2 = \alpha a_2 + \beta b_2\end{cases}$

or in matrix form,

$\begin{pmatrix}c_1\\c_2\end{pmatrix} = \begin{pmatrix}a_1&b_1\\a_2&b_2\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}$

If this system has no solution, then the coefficient matrix on the right side must be singular and its determinant would be

$\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix} = a_1b_2-a_2b_1 = 0$

and this is what we wanted to prove. QED

3 0

3 years ago

PLEASE HELP WILL MARK BRAINLIEST

Sholpan [36]

Answer: Choice D

b greater-than 3 and StartFraction 2 over 15 EndFraction

In other words,

b > 3 & 2/15

$b > 3\frac{2}{15}\\\\$

========================================================

Explanation:

Let's convert the mixed number 2 & 3/5 into an improper fraction.

We'll use the rule

a & b/c = (a*c + b)/c

In this case, a = 2, b = 3, c = 5

So,

a & b/c = (a*c + b)/c

2 & 3/5 = (2*5 + 3)/5

2 & 3/5 = (10 + 3)/5

2 & 3/5 = 13/5

The inequality $2 \frac{3}{5} < b - \frac{8}{15}\\\\$ is the same as $\frac{13}{5} < b - \frac{8}{15}\\\\$

---------------------

Let's multiply both sides by 15 to clear out the fractions

$\frac{13}{5} < b - \frac{8}{15}\\\\15*\frac{13}{5} < 15*\left(b - \frac{8}{15}\right)\\\\39 < 15b-8\\\\$

---------------------

Now isolate the variable b

$39 < 15b-8\\\\15b-8 > 39\\\\15b > 39+8\\\\15b > 47\\\\b > \frac{47}{15}\\\\b > \frac{45+2}{15}\\\\b > \frac{45}{15}+\frac{2}{15}\\\\b > 3+\frac{2}{15}\\\\b > 3\frac{2}{15}\\\\$

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how

47/15 = 3 remainder 2

The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.

7 0

3 years ago