All categories

3 years ago

Please answer these questions please

Mathematics

1 answer:

IRINA_888 [86]3 years ago

5 0

Answer:

35t 125kg

5120x12=61440 kg

10200+9=10209 kg

Step-by-step explanation:

You might be interested in

A field bordering a straight stream is to be enclosed. The side bordering the stream is not to be fenced. If 1000 yards of fenci

sveta [45]

Answer:

x = 500 yd

y = 250 yd

A(max) = 125000 yd²

Step-by-step explanation:

Let´s call x the side parallel to the stream ( only one side to be fenced )

y the other side of the rectangular area

Then the perimeter of the rectangle is p = 2*x + 2* y ( but only 1 x will be fenced)

p = x + 2*y

1000 = x + 2 * y ⇒ y = (1000 - x )/ 2

And A(r) = x * y

Are as fuction of x

A(x) = x * ( 1000 - x ) / 2

A(x) = 1000*x / 2 - x² / 2

A´(x) = 500 - 2*x/2

A´(x) = 0 500 - x = 0

x = 500 yd

To find out if this value will bring function A to a maximum value we get the second derivative

C´´(x) = -1 C´´(x) < 0 then efectevly we got a maximum at x = 500

The side y = ( 1000 - x ) / 2

y = 500/ 2

y = 250 yd

A(max) = 250 * 500

A(max) = 125000 yd²

3 0

3 years ago

Identify the expression equivalent to 16x + 24y

tatuchka [14]

Correct answer is 3: 8 (2x + 3y)

3 0

3 years ago

Read 2 more answers

Write down the 5th term in the sequence given by t(n) = n^2 +2n

madam [21]

So it our equation would be t(5)=5^2+2(5)
25+10=35 :)

5 0

3 years ago

Read 2 more answers

HELP ME I NEED THE RIGHT ANSWER

Annette [7]

Answer:

the slope is positive 10

5 0

3 years ago

One Sunday, 120 days before Christmas, Aldsworth store publishes an advertisement saying ‘120 shopping days until Christmas'. Al

Lena [83]

Answer:

(a)18

(b)1089

(c)Sunday

Step-by-step explanation:

The problem presented is an arithmetic sequence where:

First Sunday, a=1
Common Difference (Every subsequent Sunday), d=7

We want to determine the number of Sundays in the 120 days before Christmas.

(a)In an arithmetic sequence:

$\text{The nth term}, T_n=a+(n-1)d\\T_n \leq 120\\$Therefore:$\\1+7(n-1) \leq 120\\1+7n-7\leq 120\\7n-6\leq 120\\7n\leq 120+6\\7n\leq 126\\$Divide both sides by 7$\\n\leq 18$

Since the result is a whole number, there are 18 Sundays in which Aldsworth advertises.

Therefore, Aldsworth advertised 18 times.

(b)Next, we want to determine the sum of the first 18 terms of the sequence

1,8,15,...

$\text{Sum of a sequence}, S_n=\frac{n}{2}( 2a+(n-1)d)\\S_{18}=\frac{18}{2}( 2*1+(18-1)*7)\\=9(2+17*7)\\=9(2+119)\\=9*121\\S_{18}=1089$

The sum of the numbers of days published in all the advertisements is 1089.

(c)SInce the 120th day is the 18th Sunday, Christmas is on Sunday.

6 0

3 years ago

Please answer these questions please​

Please answer these questions please