How do I graph g(x)=f(x)+5

Mathematics

1 answer:

sashaice [31]3 years ago

8 0

Find the absolute value vertex. In this case, the vertex for y=|x−5|y=|x-5| is (5,0)(5,0).

(5,0)(5,0)

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

(−∞,∞)(-∞,∞)

{x|x∈R}{x|x∈ℝ}

For each xx value, there is one yy value. Select few xx values from the domain. It would be more useful to select the values so that they are around the xx value of the absolute valuevertex.

xy3241506172

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Rebecca bought groceries using 50% of her paycheck from her part-time job. She saved 75% of the remaining amount of money and ha

USPshnik [31]

0.75 (1/2x) = 60

60/3 is 20, which is 25% of her remaining money,
So her total remaining money is $80.

Then 80 x 2 = 160

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4 years ago

A confectioner has 300 pounds of a chocolate that is 1 part cocoa butter to 7 parts caramel. How much chocolate that is 1 part c

anygoal [31]

Let's call our unknown quantity $n$ . We'll need to add $n$ lbs of chocolate with a 1/15 ratio of cocoa butter to caramel to 300 lbs of chocolate with a 1/7 ratio of cocoa butter to caramel to yield $300+n$ lbs of chocolate with a 1/9 ratio of cocoa butter to caramel. We can set this up as an equation, but it's important to note something first: the ratio of the caramel to the cocoa butter is provided, but this is not the same thing as the fraction of the chocolate the cocoa butter takes up.

The 300 pounds of chocolate have 1 + 7 = 8 parts of cocoa butter and caramel total, which means that cocoa butter takes up 1/8 of those total parts; the $n$ pounds we're adding on has 1 + 15 = 16 total parts, which means the cocoa butter takes up 1/16 of those; and the $300+n$ pounds being produced have 1 + 9 = 10 total parts, so the cocoa butter takes up 1/10 of those parts. With this in mind, we can set up the following equation:

$300\big( \frac{1}{8}\big)+n\big( \frac{1}{16}\big)=(300+n)\big( \frac{1}{10}\big)$

which we can rewrite as

$\frac{300}{8}+ \frac{n}{16}= \frac{300+n}{10}$

From here, it would be helpful to combine the fractions on the left side of the equation. To do this, we'll convert the denominator of $\frac{300}{8}$ to 16, multiplying it by $\frac{2}{2}$ to obtain the fraction $\frac{600}{16}$ . Combining that with $\frac{n}{16}$ , we have:

$\frac{600+n}{16}= \frac{300+n}{10}$

To get rid of the denominator on the left, we'll multiply both sides of the equation by 16, and to eliminate the one on the right, we'll multiply both sides by 10:

$(10)(16)\big( \frac{600+n}{16}\big)=\big( \frac{300+n}{10}\big)(10)(16)$

Simplifying:

$10(600+n)=(300+n)16\\ 6000+10n=4800+16n$

And finally, we solve for $n$ :

$6000-4800=16n-10n\\ 1200=6n\\ 200=n$

So, the confectioner needs 200 lbs of chocolate that's 1 part cocoa butter and 15 parts caramel.

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4 years ago

Which expression is equivalent to 3x + 2(4y + x)? Hurry please, thanks!

docker41 [41]

Answer:

Step-by-step explanation:

4 0

3 years ago

Question 15

Ostrovityanka [42]

9514 1404 393

Answer:

A f(n)=8(0.75)^n

Step-by-step explanation:

After the first bounce, the height is 8·0.75 = 6 ft. Each subsequent bounce multiplies the previous height by 0.75, so the height after n bounces is ...

f(n) = 8·0.75^n . . . . matches choice A

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3 years ago

Enter the ordered pair for the vertices for Rx-axis(QRST).

mash [69]

Given:

The vertices of a polygon QRST are T(-2, 3), Q(1, 5), R(3, -1) and S(0, 0).

To find:

The vertices for $R_{x-axis}(QRST)$ .