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Radda [10]
3 years ago
9

Sixty photos taken at a photo shoot are in black and white. If this represents 15 percent of the total number of photos taken, w

hich equation can be used to find the total number of photos?
StartFraction 60 times 4 Over 15 times 4 EndFraction = StartFraction 240 Over 60 EndFraction
StartFraction 60 divided by 4 Over 100 divided by 4 EndFraction = StartFraction 15 Over 25 EndFraction
StartFraction 60 divided by 10 Over 100 divided by 10 EndFraction = StartFraction 6 Over 10 EndFraction
StartFraction 15 times 4 Over 100 times 4 EndFraction = StartFraction 60 Over 400 EndFraction
Mathematics
2 answers:
Oduvanchick [21]3 years ago
5 0

If this represents 15 percent of the total number of photos taken, which equation can be used to find the total number of photos? A. 60×4/15×4 = 240/60.

Lubov Fominskaja [6]3 years ago
3 0

Answer:

other guy right

Step-by-step explanation:

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625 in exponential form
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5 to the 4th power would be one way
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3 years ago
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A rectangular parking area measuring 5000 ft squared is to be enclosed on three sides using​ chain-link fencing that costs ​$5.5
kaheart [24]

Answer:

Dimensions: 75.3778 ft and 66.3325 ft

Minimum price: $1658.31

Step-by-step explanation:

Let's call the length of the parking area 'x', and the width 'y'.

Then, we can write the following equations:

-> Area of the park:

x * y = 5000

-> Price of the fences:

P = 2*x*5.5 + y*5.5 + y*7

P = 11*x + 12.5*y

From the first equation, we have that y = 5000/x

Using this value in the equation for P, we have:

P = 11*x + 12.5*5000/x = 11*x + 62500/x

To find the minimum of this function, we need to take its derivative and then make it equal to zero:

dP/dx = 11 - 62500/x^2 = 0

x^2 = 65000/11

x = 250/sqrt(11) = 75.3778 ft

This is the x value that gives the minimum cost.

Now, finding y and P, we have:

x*y = 5000

y = 5000/75.3778 = 66.3325

P = 11*x + 62500/x = $1658.31

5 0
4 years ago
The table represents an exponential function.
Stels [109]

The exponential function is given in the form

y= a (b)^{x}

where a is the initial value and b is the multiplicative rate of change

So lets plug the first two values of the table in this function , we get

2= a(b)^1

or

2= ab............................(eqn 1)

Now plug second

2/5 = a(b)^2 .............................( eqn2)

Divide equation 2 by equation 1

\frac{2=ab}{\frac{2}{5}=ab^2}

On simplifying we get

5= \frac{1}{b}

or

b=\frac{1}{5}

So the multiplicative rate of change of the function is

a) \frac{1}{5}

8 0
3 years ago
Read 2 more answers
If 2x^2-5x+7 is subtracted from x^2+2x-11, what is the leading coefficient?
guajiro [1.7K]

Answer:

-1

Step-by-step explanation:

2x² - 5x + 7 is subtracted <em>from </em>x² + 2x - 11, so

(x² + 2x - 11) - (2x² - 5x + 7)

x² + 2x - 11 - 2x² + 5x -7 (Distribute the negative)

-x² + 7x - 18 (Combine like terms)

The leading coefficient is the coefficient of the first term, so it is -1.

3 0
3 years ago
PLEASE HELP 100 POINTS!!!!!!
horrorfan [7]

Answer:

A)  See attached for graph.

B)  (-3, 0)  (0, 0)  (18, 0)

C)   (-3, 0) ∪ [3, 18)

Step-by-step explanation:

Piecewise functions have <u>multiple pieces</u> of curves/lines where each piece corresponds to its definition over an <u>interval</u>.

Given piecewise function:

g(x)=\begin{cases}x^3-9x \quad \quad \quad \quad \quad \textsf{if }x < 3\\-\log_4(x-2)+2 \quad  \textsf{if }x\geq 3\end{cases}

Therefore, the function has two definitions:

  • g(x)=x^3-9x \quad \textsf{when x is less than 3}
  • g(x)=-\log_4(x-2)+2 \quad \textsf{when x is more than or equal to 3}

<h3><u>Part A</u></h3>

When <u>graphing</u> piecewise functions:

  • Use an open circle where the value of x is <u>not included</u> in the interval.
  • Use a closed circle where the value of x is <u>included</u> in the interval.
  • Use an arrow to show that the function <u>continues indefinitely</u>.

<u>First piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

\implies g(3)=(3)^3-9(3)=0 \implies (3,0)

Place an open circle at point (3, 0).

Graph the cubic curve, adding an arrow at the other endpoint to show it continues indefinitely as x → -∞.

<u>Second piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

\implies g(3)=-\log_4(3-2)+2=2 \implies (3,2)

Place an closed circle at point (3, 2).

Graph the curve, adding an arrow at the other endpoint to show it continues indefinitely as x → ∞.

See attached for graph.

<h3><u>Part B</u></h3>

The x-intercepts are where the curve crosses the x-axis, so when y = 0.

Set the <u>first piece</u> of the function to zero and solve for x:

\begin{aligned}g(x) & = 0\\\implies x^3-9x & = 0\\x(x^2-9) & = 0\\\\\implies x^2-9 & = 0 \quad \quad \quad \implies x=0\\x^2 & = 9\\\ x & = \pm 3\end{aligned}

Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.

Set the <u>second piece</u> to zero and solve for x:

\begin{aligned}\implies g(x) & =0\\-\log_4(x-2)+2 & =0\\\log_4(x-2) & =2\end{aligned}

\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b

\begin{aligned}\implies 4^2&=x-2\\x & = 16+2\\x & = 18 \end{aligned}

Therefore, the x-intercept for the second piece is (18, 0).

So the x-intercepts for the piecewise function are (-3, 0), (0, 0) and (18, 0).

<h3><u>Part C</u></h3>

From the graph from part A, and the calculated x-intercepts from part B, the function g(x) is positive between the intervals -3 < x < 0 and 3 ≤ x < 18.

Interval notation:  (-3, 0) ∪ [3, 18)

Learn more about piecewise functions here:

brainly.com/question/11562909

3 0
2 years ago
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