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3 years ago

14

binks need to raise at least $3600. He was given $200 and plans to save $50 per week. Select all the inequalities in this situat

ion where , n, is the number of weeks.

1 answer:

-BARSIC- [3]3 years ago

6 0

Answer:

First of read all formulas

Step-by-step explanation:

Calculate 50 *7 and add 350 and 200

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3 years ago

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Enter the value of h(-4) for h(x)=(x-2)2+3x in the box.

-Dominant- [34]

Step-by-step explanation:

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2 years ago

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On a day, a video was posted online, 5 people watched the video. The next day the number of viewers had doubled. This tend conti

aleksklad [387]

Answer:

Day 13

Step-by-step explanation:

The question can be represented with a geometric sequence

Where,

a = 5

Common ratio , r = 2

v(t) = a * r^(t - 1)

v(t) = 5 * 2^(t-1)

20,000 = 5 * 2^(t-1)

Divide both sides by 5

20,000 / 5 = 5 * 2^(t-1) / 5

4,000 = 2^(t-1)

2^12 = 2^(t-1)

12 = t - 1

12 + 1 = t

t = 13 days

What would be the day when more than 20,000 people will see the video

Check:

v(t) = 5 * 2^(t-1)

= 5 * 2^(13-1)

= 5 * 2^12

= 5 * 4,096

= 20,480

Therefore,

The day when more than 20,000 people will see the video is day 13

6 0

3 years ago

Solve: x^+4/x-1 = 5/x-1

AlexFokin [52]

The first step for solving this equation is to determine the defined range.
$\frac{ x^{4} }{x-1} = \frac{5}{x-1}$ , x ≠ 1
Remember that when the denominators of both fractions are the same,, you need to set the numerators equal. This will look like the following:
$x^{4}$ = 5
Take the root of both sides of the equation and remember to use both positive and negative roots.
x +/- $\sqrt[4]{5}$
Separate the solutions.
x = $\sqrt[4]{5}$ , x ≠ 1
x = - $\sqrt[4]{5}$
Check if the solution is in the defined range.
x = $\sqrt[4]{5}$
x = - $\sqrt[4]{5}$
This means that the final solution to your question are the following:
x = $\sqrt[4]{5}$
x = - $\sqrt[4]{5}$
Let me know if you have any further questions.
:)

7 0

3 years ago

Which piece wise function is graphed below

AlekseyPX

Answer:

Option A

Step-by-step explanation:

Here is how to approach the problem:

We see that all our restrictions for all four answer choices are relatively the same with a couple of changes here and there.
One way to eliminate choices would be to look at which restrictions don't match the graph.

At x<-5, there is a linear function that does have a -2 slope and will intersect the x axis at -7. The line ends with an open circle, so any answer choice with a linear restriction of x less than or equal to -5 is wrong. This cancels out choices C and D.

Now we have two choices left.

For the quadratic in the middle, the vertex is at (-2,6) and the vertex is a maximum, meaning our graph needs to have a negative sign in front of the highest degree term. In our case, none of our quadratics left are in standard form, and instead are in vertex form.

Vertext form is f(x) = a(x-h)^2 + k.

h being the x-coordinate of the vertex and k being the y-coordinate.

We know that the opposite of h will be the actual x-coordinate of the vertex, so if our vertex is -2, we will see x+2 inside the parenthesis. This leaves option A as the only correct choice.

3 0

2 years ago