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Levart [38]
2 years ago
5

What is the vertex of the graph of f(x) = |x 3| 7? (3, 7) (7, 3) (–3, 7) (7, –3).

Mathematics
2 answers:
KatRina [158]2 years ago
5 0

The vertex of the given modulus function is (3,-7).

Given information:

The given modulus function is,

f(x)=|x-3|-7

It is required to find the vertex of the given function.

<h3>What is the vertex of a modulus function?</h3>

Modulus function is in the shape of V. The notch of V is the vertex of the function.

The given function can be defined as,

f(x)=x-10;x\geq3\\&#10;f(x)=-x-4;x

From the above function, it can be concluded that the vertex of the function should be (3,-7). Also shown in the attached image.

See the attached image.

Therefore, the vertex of the given modulus function is (3,-7).

For more details about modulus function, refer to the link:

brainly.com/question/18793028

marshall27 [118]2 years ago
5 0

The vertex of the graph of f(x) is (-3, 7).

<h2>Given that</h2>

Graph; \rm  f(x) = |x + 3| + 7

<h3>We have to determine</h3>

What is the vertex of the graph of f(x)?

According to the question

The standard form of the absolute value function is;

\rm y = a|x-h|+k

Where h and k are the vertexes of the function.

Graph; \rm  f(x) = |x + 3| + 7

Converting the equation into the standard form the absolute value function;

\rm  f(X)=|x + 3| + 7\\&#10;\\&#10;f(x) = |x-(-3)|+7

On comparing with the standard absolute value function the vertices of the graph f(x).

h = -3 and k = 7

Hence, the vertex of the graph of f(x) is (-3, 7).

To know more about Absolute Function click the link given below.

brainly.com/question/1883473

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If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
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If k is odd, then

\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor

while if k is even, then the sum would be

\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4

The latter case is easier to solve:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

which means k=6.

In the odd case, instead of considering the above equation we can consider the partial sums. If k is odd, then the sum of the even integers between 1 and k would be

S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)

Now consider the partial sum up to the second-to-last term,

S^*=2+4+6+\cdots+(k-5)+(k-3)

Subtracting this from the previous partial sum, we have

S-S^*=k-1

We're given that the sums must add to 2k, which means

S=2k
S^*=2(k-2)

But taking the differences now yields

S-S^*=2k-2(k-2)=4

and there is only one k for which k-1=4; namely, k=5. However, the sum of the even integers between 1 and 5 is 2+4=6, whereas 2k=10\neq6. So there are no solutions to this over the odd integers.
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melamori03 [73]
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