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2 years ago

1. 7p - 6pc + 3c - 2 Number of terms: Coefficients: Constant terms:

Mathematics

1 answer:

Andreas93 [3]2 years ago

8 0

Step-by-step explanation:

Given:

7p - 6pc + 3c - 2

Find:

Number of terms

Coefficients

Constant terms

Computation:

Number of terms = 4

Coefficients = (7, -6, 3)

Constant terms = -2

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Plz help I need help

fenix001 [56]

Answer:

3 / 4 * 50 = 150 / 4

= 37.5 or 37½ yards

Step-by-step explanation:

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2 years ago

In order to have 290,000 in 10 yrs you should deposit how much each month

never [62]

Answer:2416{APPROX}

Step-by-step explanation:

AMOUNT=290000

NO OF YEARS=10

NO OF MONTHS IN 10 YEARS=120 MONTHS.

SO AMOUNT TO DEPOSIT PER MONTH=290000/120=2416(APPROX)

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2 years ago

it takes someone 15 minutes to prepare 3 1/4 cups of juice how many hours would it take to prepare 32 1/2 cups of juice

wariber [46]

$\bf \begin{array}{ccll} minutes&\stackrel{juice}{cups}\\ \cline{1-2} 15&3\frac{1}{4}\\\\ x&32\frac{1}{2} \end{array}\implies \cfrac{15}{x}=\cfrac{3\frac{1}{4}}{32\frac{1}{2}}\implies \cfrac{15}{x}=\cfrac{\frac{3\cdot 4+1}{4}}{\frac{32\cdot 2+1}{2}}\implies \cfrac{15}{x}=\cfrac{\frac{13}{4}}{\frac{65}{2}}$

$\bf \cfrac{15}{x}=\cfrac{~~\begin{matrix} 13 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{\underset{2}{~~\begin{matrix} 4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\cdot \cfrac{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{\underset{5}{~~\begin{matrix} 65 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\implies \cfrac{15}{x}=\cfrac{1}{10}\implies 150=x\leftarrow \begin{array}{llll} \textit{150 minutes or}\\\\ \textit{2 hours and a half} \end{array}$

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2 years ago

What is the product of (y+3)(y^2- 3y+9)

makvit [3.9K]

Answer:

y^3+27

Step-by-step explanation:

$(y+3)(y^2-3y+9)\\$

*use box method

3 0

2 years ago

B. Determine the values of a and b so that f is continuous.<br><br> Please help!

wlad13 [49]

Answer:

following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

$\lim_{x\to 2}+f(x) = \lim_{x\to 2}+ (ax^2-bx+3) = 4a-2b+3\\\\ \to 4a-2b+3 = 4\\\\ \therefore \ \ 4a-2b = 1\\\\$

The one-sided limits of F(x) at x = 3 must be equivalent for f(x) to be continuous at x = 3.

$\to \lim_{x\to 3}- f(x) = \lim_{x\to 3}- (ax^2-bx+3) = 9a-3b+3\\\\ \to \lim_{x\to 3}+ f(x) = \lim_{x\to 3}+ (2x-a+b) = 6-a+b\\\\$

So,

$\to 9a-3b+3 = 6-a+b\\\\\therefore\\ \to 10a -4b = 3$

$\to 4a - 2b = 1......(a)\\\to 10a - 4b = 3.......(b)\\$

In equation a multiply the by -2 and then add in the equation b:

$\to -8a + 4b = -2\\\to 10a - 4b = 3\\ \to 2a = 1\\\\ \to a = \frac{1}{2}\\\\ \to \ 4(\frac{1}{2}) - 2b = 1\\\\ \to 2 - 2b = 1\\\\ \to -2b = -1\\\\ \to b = \frac{1}{2}$

So, the value of $a \ and \ b= \frac{1}{2}$

4 0

2 years ago

1. 7p - 6pc + 3c - 2 Number of terms: Coefficients: Constant terms:​

1. 7p - 6pc + 3c - 2 Number of terms: Coefficients: Constant terms: