All categories

3 years ago

Functions help?

Mathematics

1 answer:

Tcecarenko [31]3 years ago

6 0

Question A is a linear graph because it graphs as a diagonal line, while on the other hand Question B graphs as exponential! Hope this helped!

You might be interested in

What is 1 hundredth + 3 tenths=

musickatia [10]

Answer is 0.31. Hope this helps.

6 0

3 years ago

Solve the following question algebraically. SOMEONE PLEASE HELP ME! ILL MARK THE BRAINLIEST ANSWER!!I’VE TRIED TO FIGURE IT OUT

Delicious77 [7]

Solve for x. Isolate the x. Note the equal sign. What you do to one side, you do to the other. Do the opposite of PEMDAS.

First, multiply 3 to both sides

6(3) = ((x + 2)/3)(3)

18 = x + 2

Finally, isolate the x. Subtract 2 from both sides

18 (-2) = x + 2 (-2)

x = 18 - 2

x = 16

16 is your answer for x

hope this helps

8 0

3 years ago

If f(x)=3|x-2|, what is f(5.9)? a: 9 b: 10 c: 11 d: 12

Mice21 [21]

$|a|= \left\{\begin{array}{ccc}a&if&a\geq0\\-a&if&a \ \textless \ 0\end{array}\right\\\\f(x)=3|x-2|\\\\f(5.9)=3|5.9-2|=3|3.9|=3\cdot3.9=11.7$

7 0

3 years ago

Read 2 more answers

5^2(7) <br> Pls solve I am stuck

AfilCa [17]

Answer:

$5^{2}$ (7) = 175

Step-by-step explanation:

hope this helps! just follow the order of operations. :)

3 0

3 years ago

A fence is to be built to enclose a rectangular area of 1800 square feet. The fence along three sides is to be made of material

drek231 [11]

Answer:

length = 60 foot, width = 30 foot

Step-by-step explanation:

Area of rectangular part, A = 1800 ft²

Cost of fencing three sides is $ 6 per foot and cost of one side fencing is $18 per foot.

Let the length of the rectangle is l and the width of the rectangle is W.

Area = Length x width

A = L x W

1800 = L x W ...... (1)

Total cost of fencing, C = 6 x ( L + W + L) + 18 x W

C = 6 (2L + W) + 18 W

C = 12 L + 24 W

Substitute the value of W from equation (1), $W=\frac{1800}{L}$ in equation (2)

$C=12L+24\times \frac{1800}{L}$

$C=12L+\frac{43200}{L}$

Differentiate both sides with respect to L:

$\frac{dC}{dL}=12-\frac{43200}{L^{2}}$

Put it equal to zero for maxima and minima

$12-\frac{43200}{L^{2}}=0$

L = 60 foot

and W = 30 foot

So, the costing is minimum for length = 60 foot and the width = 30 foot.

4 0

3 years ago