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3 years ago

Which relations are functions and which are not functions?

1 answer:

3 0

If the x value is repeated is is not a function. (unless it is a negative and a positive)
So lets see:
For A we have 3 6 9 12 no repeations so therefore It is A
For B we have 2 2 3 9, 2 is repeated so therefore it is B
For C we have 4 6 4 6, both 4 and 6 were repeated so therefore is is B
and lastly for D we have 3 5 6 6
And 6 was repeated so therefore it is B.
A)A
B)B
C)B
D)B

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The value V of a certain automobile that is t years old can be modeled by V(t) = 14,651(0.81). According to the model, when will

lakkis [162]

Answer:

a) The car will be worth $8000 after 2.9 years.

b) The car will be worth $6000 after 4.2 years.

c) The car will be worth $1000 after 12.7 years.

Step-by-step explanation:

The value of the car after t years is given by:

$V(t) = 14651(0.81)^{t}$

According to the model, when will the car be worth V(t)?

We have to find t for the given value of V(t). So

$V(t) = 14651(0.81)^{t}$

$(0.81)^t = \frac{V(t)}{14651}$

$\log{(0.81)^{t}} = \log{(\frac{V(t)}{14651})}$

$t\log{(0.81)} = \log{(\frac{V(t)}{14651})}$

$t = \frac{\log{(\frac{V(t)}{14651})}}{\log{0.81}}$

(a) $8000

V(t) = 8000

$t = \frac{\log{(\frac{8000}{14651})}}{\log{0.81}} = 2.9$

The car will be worth $8000 after 2.9 years.

(b) $6000

V(t) = 6000

$t = \frac{\log{(\frac{6000}{14651})}}{\log{0.81}} = 4.2$

The car will be worth $6000 after 4.2 years.

(c) $1000

V(t) = 1000

$t = \frac{\log{(\frac{1000}{14651})}}{\log{0.81}} = 12.7$

The car will be worth $1000 after 12.7 years.

5 0

2 years ago

How are exponential modelsare helpful in estimating future values of a function?

alexandr1967 [171]

I think it’s 17.4 idk but hope it’s helpful

3 0

2 years ago

ASAP!! Please help me. I will not accept nonsense answers, but will mark as BRAINLIEST if you answer is correctly with solutions

aliina [53]

Answer: B. H(t) = -6.9(t - 2.3)² + 112

<u>Step-by-step explanation:</u>

The vertex form of a quadratic equation is: y = a(x - h)² + k

where (h, k) is the vertex ⇒ k is the maximum height

the distance traveled is when y = 0 ⇒ $x=\sqrt{\dfrac{-k}{a}}+h$

Given: H(t) = -7.1(t - 2.3)² + 98

maximum height (k) = 98 feet

distance traveled (x) = $\sqrt{\dfrac{-98}{-7.1}}+2.3$ = 6.02 seconds

A) H(t) = -7.5(t - 2.2)² + 112

maximum height (k) = 112 feet

distance traveled (x) = $\sqrt{\dfrac{-112}{-7.5}}+2.2$ = 6.06 seconds

B) H(t) = -6.9(t - 2.3)² + 112

maximum height (k) = 112 feet

distance traveled (x) = $\sqrt{\dfrac{-112}{-6.9}}+2.3$ = 6.33 seconds

C) H(t) = -6.9(t - 2.4)² + 95

maximum height (k) = 95 feet

This has a lower height than the given equation.

D) H(t) = -7.5(t - 2.3)² + 95

maximum height (k) = 95 feet

This has a lower height than the given equation.

Both options A and B travel higher and stay in the air longer than last year's winner, however option B stays in the air longer than option A.

5 0

2 years ago

Five years ago, the tree you planted measured 4 meters 9 decimeters tall. Today that tree measures 6 meters, 1 decimeter. How mu

Gnesinka [82]

Answer: 1.2metres

Step-by-step explanation:

Five years ago, tree was 4.9m

Today, tree is 6.1m

Subtract 4.9m from 6.1m

Tree has grown by 1.2m

7 0

3 years ago

If you answer this, i might give you 3 extra points (:

rodikova [14]

Jewelery maker has
24 jade beads
30 teak Beads
So we need to find the highest common factor for both 24 and 30
The factors for both numbers are
24 -1,2,3,4,6,8,12,24
30 - 1,2,3,5,6,10,15,30
The highest common factor for both is 6
The greatest number of necklaces she can make is 6
Number of jade beads - 24/6 = 4 jade beads
Number of teak beads - 30/6 = 5 teak beads
So each necklace will have 4 jade beads and 5 teak beads

7 0

3 years ago

Read 2 more answers