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2 years ago

Someone please help me!

Mathematics

1 answer:

stiv31 [10]2 years ago

5 0

My answer is probably 40+40 or 40×40or 40÷40 cause 40 times 40 equals might equal 1,600 or1,000 40 plus 40 equals 80 tho so that's my best guess 40 divided by 40 is probably just the same as 40 times 40 1,000 or 1,600.

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How would i graph y=− 5/2x-3 and x−y=−4 (_,_)

SashulF [63]

Given:

The system of equations is

$y=-\dfrac{5}{2}x-3$

$x-y=-4$

To find:

The graph of the given system of equations and its solution.

Solution:

The table of values for $y=-\dfrac{5}{2}x-3$ is

x y

-4 7

-2 2

0 -3

Plot these points and connect them by a straight line.

The table of values for $x-y=-4$ is

x y

-4 0

-2 2

0 4

Plot these points and connect them by a straight line.

From the below graph it is clear that the lines intersect each other at point (-2,2).

Therefore, the solution of given system of equation is (-2,2).

6 0

3 years ago

Which whole number is closest to the value of square root of 54?

Blababa [14]

The answer for this problem is 7

3 0

3 years ago

Read 2 more answers

Each week, Heather’s company has $5000 in fixed costs plus an additional $250 for each system produced. The company is able to p

kvv77 [185]

The question is an illustration of composite functions.

Functions c(n) and h(n) are $\mathbf{c(n) = 5000 + 250n}$ and $\mathbf{n(h) = 5h}$
The composite function c(n(h)) is $\mathbf{c(n(h)) = 5000 + 1250h}$
The value of c(n(100)) is $\mathbf{c(n(100)) = 130000}$
The interpretation is: "the cost of working for 100 hours is $130000"

The given parameters are:

$5000 in fixed costs plus an additional $250
5 systems in one hour of production

(a) Functions c(n) and n(h)

Let the number of system be n, and h be the number of hours

So, the cost function (c(n)) is:

$\mathbf{c(n) = Fixed + Additional \times n}$

This gives

$\mathbf{c(n) = 5000 + 250 \times n}$

$\mathbf{c(n) = 5000 + 250n}$

The function for number of systems is:

$\mathbf{n(h) = 5 \times h}$

$\mathbf{n(h) = 5h}$

(b) Function c(n(h))

In (a), we have:

$\mathbf{c(n) = 5000 + 250n}$

$\mathbf{n(h) = 5h}$

Substitute n(h) for n in $\mathbf{c(n) = 5000 + 250n}$

$\mathbf{c(n(h)) = 5000 + 250n(h)}$

Substitute $\mathbf{n(h) = 5h}$

$\mathbf{c(n(h)) = 5000 + 250 \times 5h}$

$\mathbf{c(n(h)) = 5000 + 1250h}$

(c) Find c(n(100))

c(n(100)) means that h = 100.

So, we have:

$\mathbf{c(n(100)) = 5000 + 1250 \times 100}$

$\mathbf{c(n(100)) = 5000 + 125000}$

$\mathbf{c(n(100)) = 130000}$

(d) Interpret (c)

In (c), we have: $\mathbf{c(n(100)) = 130000}$

It means that:

The cost of working for 100 hours is $130000

Read more about composite functions at:

brainly.com/question/10830110

5 0

3 years ago

Find the difference.

Vsevolod [243]

$5.\\12\dfrac{3}{10}-7\dfrac{7}{10}=11\dfrac{10+3}{10}-7\dfrac{7}{10}=11\dfrac{13}{10}-7\dfrac{7}{10}=4\dfrac{6}{10}=4\dfrac{6:2}{10:2}=\boxed{5\dfrac{3}{5}}\\\\11-7=4\\\\\dfrac{13}{10}-\dfrac{7}{10}=\dfrac{13-7}{10}=\dfrac{6}{10}\\\\6.\\8\dfrac{1}{6}-3\dfrac{5}{6}=7\dfrac{6+1}{6}-3\dfrac{5}{6}=7\dfrac{7}{6}-3\dfrac{5}{6}=(7-3)+\dfrac{7-5}{6}=4\dfrac{2}{6}=4\dfrac{2:2}{6:2}=\boxed{4\dfrac{1}{3}}$

$9.\\7\dfrac{1}{6}-2\dfrac{5}{6}=6\dfrac{6+1}{6}-2\dfrac{5}{6}=6\dfrac{7}{6}-2\dfrac{5}{6}=(6-2)+\dfrac{7-5}{6}=4\dfrac{2}{6}=\boxed{4\dfrac{1}{3}}\\\\10.\\9\dfrac{3}{12}-4\dfrac{7}{12}=8\dfrac{12+3}{12}-4\dfrac{7}{12}=8\dfrac{15}{12}-4\dfrac{7}{12}=(8-4)+\dfrac{15-7}{12}=4\dfrac{8}{12}\\\\=4\dfrac{8:4}{12:4}=\boxed{4\dfrac{2}{3}}$

5 0

3 years ago

HELP WILL GIVE BRAINLIST PLEASE! slope (m) = ? y-intercept (b) = ?

timama [110]

Answer: slope=5 y-intercept=-1

Step-by-step explanation:

Slope=rise/run

In this case we rise 5 and run 1

5/1=5

On the graph you can see the line passes through the y axis at -1

5 0

3 years ago