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2 years ago

Please help!

Mathematics

2 answers:

lions [1.4K]2 years ago

3 0

Michelle's result shows f(2) = -13, selection C.

chubhunter [2.5K]2 years ago

3 0

Answer:

C. -13

Step-by-step explanation:

The polynomial f(x) she is dividing would be f(x)=1x²-7x-3. This is because in synthetic division, the numbers used are the coefficients of the polynomial, and the polynomial will have a degree one larger than the divisor. Since the divisor is x-2, with a degree of 1, the polynomial will have a degree of 2, meaning it begins with an x².

The first coefficient is 1; this is 1x². The second coefficient is -7; this is -7x. The last coefficient is -3; this is -3. Together, this gives us

f(x) = 1x²-7x-3

To find f(2), substitute 2 in for every x:

f(2) = 1(2²)-7(2)-3 = 1(4)-14-3 = 4-14-3 = -10-3 = -13

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Galina-37 [17]

Hey there :)

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2 years ago

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AveGali [126]

Answer:

$\boxed{\bold{\huge{\boxed{Area\ of\ Trapezium = 150\ cm\²}}}}$

Step-by-step explanation:

Using Pythagorean Theorem first to find EC

$\sf CB^2 = EB^2+EC^2\\Where \ CB = 13 \ and \ EB = 5 \\13^2 = 5^2 + EC^2 \\169 - 25 = EC^2 \\EC^2 = 144\\Taking \ sqrt \ on \ both \ sides\\EC = 12 \ cm$

Given that:

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Where AB = 17 , DC = 8 and EC = 12

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2 years ago

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inna [77]

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2 years ago

Order -3.5 ℉ 5 ℉ 1.5 ℉ -0.5 ℉ -2℉, 2.5℉ -4 ℉ from least to greatest

Levart [38]

Answer:

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Step-by-step explanation:

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2 years ago

How to solve number 13

MakcuM [25]

a) Each year the price increases by 3%, so if burger initially has price $p=\$3.99$ , then next year its price will be $1.03p\approx\$4.11$ . After another year, this price will increase by 3%, so that the burger would cost $1.03^2\approx\$4.23$ .

b) If the same pattern continues, then 10 years later we should expect the price to be $1.03^{10}p\approx\$5.36$ .

c and d) The pattern appears to be that the price $p$ of the burger after $t$ years, starting at a price $p_0$ , is

$p(t)=1.03^tp_0$

which forms a geometric sequence that diverges. This necessarily means that the price would eventually exceed $20, and we find that that time comes about

$\$20=1.03^t(\$3.99)\implies t\approx54.5$

years later.

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2 years ago