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2 years ago

11

A 3rd degree binomial with a constant term of 8

2 answers:

rewona [7]2 years ago

7 0

Answer:

-5x^3 + 8

Step-by-step explanation:

ZanzabumX [31]2 years ago

6 0

Answer:

-5x^3 + 8

Step-by-step explanation:

ive done this problem in khan academy before

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Please i need the answer

Harrizon [31]

Answer: b
sorry cant give explanation

7 0

2 years ago

PLEASE HELP I NEED THIS ASAP<br> Find the vertex of y=x^2-7x+4

Goshia [24]

Answer:

$\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)$

Step-by-step explanation:

$y=x^2-7x+4\\\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}\\\mathrm{The\:parabola\:params\:are:}\\a=1,\:b=-7,\:c=4\\x_v=-\frac{b}{2a}\\x_v=-\frac{\left(-7\right)}{2\cdot \:1}\\\mathrm{Simplify}\\x_v=\frac{7}{2}\\\mathrm{Plug\:in}\:\:x_v=\frac{7}{2}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}\\y_v=\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4\\$

$\mathrm{Simplify\:}\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4:\quad -\frac{33}{4}\\y_v=-\frac{33}{4}\\Therefore\:the\:parabola\:vertex\:is\\\left(\frac{7}{2},\:-\frac{33}{4}\right)\\\mathrm{If}\:a0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}\\a=1\\\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)$

6 0

3 years ago

What is the exponent of x in in the expression -4x^3y^2 + 6

liberstina [14]

Answer:

3

Step-by-step explanation:

- 4x³y² + 6

the x is raised to the power of 3 , that is exponent of x is 3

4 0

2 years ago

The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. a) What is the probab

LiRa [457]

Answer:

a) 0.0082

b) 0.9987

c) 0.9192

d) 0.5000

e) 1

Step-by-step explanation:

The question is concerned with the mean of a sample.

From the central limit theorem we have the formula:

$z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$

a) $z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40$

The area to the left of z=2.40 is 0.9918

The area to the right of z=2.40 is 1-0.9918=0.0082

$\therefore P(\bar X\:>\:1224)=0.0082$

b) $z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00$

The area to the left of z=3.00 is 0.9987

$\therefore P(\bar X\:$

c) The z-value of 1200 is 0

The area to the left of 0 is 0.5

$z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40$

The area to the left of z=1.40 is 0.9192

The probability that the sample mean is between 1200 and 1214 is

$P(1200\:$

d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000

e) $z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65$

The area to the left of z=-112.65 is 0.

The area to the right of z=-112.65 is 1-0=1

5 0

2 years ago

Pls helppp it's due after 30 min

Sergio [31]

Step-by-step explanation:

<h3>• C. P = Cost Price.. , S. P (Sell Price) , </h3>

<h3>→ C.P = 100×S.P/100+Profit (when profit) </h3>

<h3>(a) C.P = 100 × 55/100 + 10 = 5500/110 </h3>

<h2> C.P » $ 50</h2><h3 /><h3 /><h3>(b) C.P = 100 × 558/100+24 = 55800/124 </h3>

<h2> C.P » $ 450</h2><h3 /><h3 /><h3 /><h3 /><h3 /><h3>→ C. P = 100 × S.P/100 - loss (when loss) </h3>

<h3> (c) C. P = 100 × 680 / 100 - 15 = 68000/85</h3>

<h2> C. P » $ 800</h2>

<h3>(d) C. P = 100 × 11.78 / 100 - 5 = 1178/95 </h3>

<h2> C. P » $ 12.4</h2>

4 0

2 years ago