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Tpy6a [65]
3 years ago
5

Which table shows y as a function of x?

Mathematics
2 answers:
Finger [1]3 years ago
7 0

Answer:

Top one

Step-by-step explanation:

If it is not possible to draw a vertical line to touch the graph of a function in more than one place, then y is a function of x.

Sonbull [250]3 years ago
6 0

Answer: The first one

Brainliest pls if it is correct!

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Determine the intervals on which the function is increasing, decreasing, and constant. (5 points)
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The one above the one you have now
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ella [17]
If she has 6 yards and each pillow required 1 yard she can make 6 pillows
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Given that P = (-7, 16) and Q = (-8, 7), find the component form and magnitude of vector QP .
Katyanochek1 [597]

Answer:

a)The component form is

\vec {QP}=\binom{ - 1}{ - 8}

b)The magnitude is √65

c) <2,14>

Step-by-step explanation:

Recall that:

\vec {QP}=\vec {OP}-\vec{OQ}

We substitute the position vectors to get:

\vec {QP}=\binom{ - 8}{7} -  \binom { - 7}{15}

We subtract the corresponding components to obtain:

\vec {QP}=\binom{ - 8 -  - 7}{7 - 15}

This gives:

\vec {QP}=\binom{ - 8  + 7}{7 - 15}

This simplifies to:

\vec {QP}=\binom{ - 1}{ - 8}

The magnitude of a vector in the component form:

\binom{x}{y}

is

\sqrt{ {x}^{2}  +  {y}^{2} }

This means:

|\vec {QP}|= \sqrt{ {( - 1)}^{2}  +  {( - 8)}^{2} }

This simplifies to:

|\vec {QP}| = \sqrt{ 1 +  64 }

|\vec {QP}| = \sqrt{ 65 }

c) We have the vectors u = <4, 8>, v = <-2, 6>.

We want to find:

u+v

This implies that:

u+v=<4,8>+<-2,6>

We add the corresponding components to get;

u+v=<4+-2,8+6>

This simplifies to:

u+v=<2,14>

7 0
3 years ago
each of exercises 15–30 gives a function ƒ(x) and numbers l, c, and e 7 0. in each case, find an open interval about c on which
3241004551 [841]

The given inequality holds for the open interval (2.97,3.03)

It is given that

f(x)=6x+7

cL=25

c=3

ε=0.18

We have,

|f(x)−L| = |6x+7−25|

          = |6x−18|

          = |6(x−3)|

          = 6|x−3|

Now,

6|x−3| <0.18  then |x−3|<0.03 ----->−0.03<x-3<0.03---->2.97<x<3.03

the given inequality holds for the open interval (2.97,3.03)

For more information on inequality click on the link below:

brainly.com/question/11613554

#SPJ4

Although part of your question is missing, you might be referring to this full question: For the given function f(x) and values of L,c, and ϵ0, find the largest open interval about c on which the inequality |f(x)−L|<ϵ holds. Then determine the largest value for δ>0 such that 0<|x−c|<δ→|f(x)−|<ϵ.

f(x)=6x+7,L=25,c=3,ϵ=0.18

 

.

6 0
2 years ago
PLEASE HELP!!!!!
seraphim [82]

Answer:

2,5,6,15

Step-by-step explanation:

that's what I think

6 0
3 years ago
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