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Vikki [24]
3 years ago
3

Which statements can be used to compare the functions? Select two options. Only f(x) has an x-intercept. g(x) has the greatest m

aximum. All three functions have a y-intercept. All three functions have the same range values. All three functions have the same domain values.
Mathematics
1 answer:
Levart [38]3 years ago
3 0

Answer:

Only f(x) has an x-intercept.

All three functions have the same domain values.

Step-by-step explanation:

Given

See attachment for table

Required

Select two correct options

<u>Checking the domain:</u>

The x values represent the domain.

From the attached table, we have:

x = \{-4.-3,-2,-1,1\}

These values of x apply to the three functions; f(x), g(x) and h(x).

This means that the functions have the same domain

<u>Checking the greatest maximum</u>

The maximum of each function is:

f(x) = 1

g(x) = -4

h(x) = 2

Compare these three values, we have:

h(x) > f(x) > g(x)

This means that h(x) has the greatest maximum

<u>Checking the x intercept</u>

When a function equals 0, then it has an x intercept

Of the three functions, we have:

f(-3) = 0 --- This represents the x intercept.

Other functions do not have this value.

This means that only f(x) has an x intercept

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Franky puts £1000 in a bank account for six years with 2.5% rate of interest how much will be in the account at the end of that
yanalaym [24]

Answer:

£1159.69

Step-by-step explanation:

Assuming that the rate of interest is compound interest, you perform the equation:

New amount = Original amount × InterestRate^Number of Years

New amount = 1000 × 1.025⁶

New amount = 1159.693418

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8 0
3 years ago
$6,000 dollars is invested in two different accounts earning 3% and 5% interest. at the end of one year, the two accounts earned
Anettt [7]

<u>Answer-</u>

<em>$2000</em><em> were invested at 5%</em>

<u>Solution-</u>

Let x amount of money was invested at 5% and (6000-x) amount was invested as 3%

We know that,

i=\dfrac{P\cdot r\cdot t}{100}

Putting the values,

i_1=\dfrac{P_1\cdot r_1\cdot t}{100}

i_1=\dfrac{x\cdot 5\cdot 1}{100}=\dfrac{5x}{100}

And

i_2=\dfrac{P_2\cdot r_2\cdot t}{100}

i_2=\dfrac{(6000-x)\cdot 3\cdot 1}{100}=\dfrac{3(6000-x)}{100}

According to the question,

\Rightarrow \dfrac{5x}{100}+\dfrac{3(6000-x)}{100}=220

\Rightarrow \dfrac{5x+3(6000-x)}{100}=220

\Rightarrow 5x+3(6000-x)=22000

\Rightarrow 5x+18000-3x=22000

\Rightarrow 5x-3x=22000-18000

\Rightarrow 2x=4000

\Rightarrow x=2000

Therefore, $2000 were invested at 5%

8 0
4 years ago
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Step-by-step explanation:

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