3429111 what is the place value of the 4

Which graph represents the function f(x) = |x + 3|?

Sophie [7]

Answer: (B)
Explanation: If you are unsure about where to start, you could always plot some numbers down until you see a general pattern.

But a more intuitive way is to determine what happens during each transformation.
A regular y = |x| will have its vertex at the origin, because nothing is changed for a y = |x| graph. We have a ray that is reflected at the origin about the y-axis.

Now, let's explore the different transformations for an absolute value graph by taking a y = |x + h| graph.
What happens to the graph?
Well, we have shifted the graph -h units, just like a normal trigonometric, linear, or even parabolic graph. That is, we have shifted the graph h units to its negative side (to the left).

What about the y = |x| + h graph?
Well, like a parabola, we shift it h units upwards, and if h is negative, we shift it h units downwards.

So, if you understand what each transformation does, then you would be able to identify the changes in the shape's location.

6 0

3 years ago

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The product of 3.41 and 8.5 will have how many decimal places?

Aleks04 [339]

Hello!

3.41 × 8.5 = 28.985

The product of 3.41 and 8.5 is 28.985, which has three decimal places.

3 0

3 years ago

In right triangle ABC with right angle C, AC=9 and the measure of angle B=30 degree. Find the length of AB.

pshichka [43]

$From\ triangle\ with\ angles: 30^{\circ},60^{\circ},90^{\circ}\\\
|AB|=a\sqrt{3}\\
|AC|=a=9\\\\
|AB|=9\sqrt{3}\\\\
Length\ of\ |AB|\ is\ equal\ to\ 9\sqrt{3}.
$

3 0

2 years ago

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Jack's mother gave him 50 chocolates to give to his friends at his birthday party. He gave 3 chocolates to each of his friends a

djverab [1.8K]

50 = 3x + 2

50 - 2 = 3x

48 = 3x

48/3 = 3x/3

16 = x

Jack has 16 friends.

4 0

2 years ago

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(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago