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4 years ago

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On Monday joe bought 10 cups of coffee and 5 doughnuts for his office at the cost of $16.50. On Tuesday he bought 5 cups of coff

ee and 10 doughnuts for a total of 14.25. How much was each cup of coffee

1 answer:

svp [43]4 years ago

4 0

Answer:

each cup cost 1.25 on Tuesday

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Which is the following equation rewritten in slope-intercept form: -8x=2-2y. A) y=1/4x+1. B) y=4x+1. C). Y=-4x+1. D) y=-4x-1

nevsk [136]

Step-by-step explanation:

-8x = 2 - 2y

2y - 8x = 2

2y = 8x + 2

y = 4x + 1

Answer B.

8 0

3 years ago

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Here are the first four terms of a quadratic sequence.

Llana [10]

Answer:

a = 3, b = 1, c = - 3

Step-by-step explanation:

Substitute n = 1, 2, 3 into the n th term

a + b + c = 1 → (1)

4a + 2b + c = 11 → (2)

9a + 3b + c = 27 → (3)

Subtract (1) from (2) term by term to eliminate c

Subtract (2) from (3) term by term to eliminate c

3a + b = 10 → (4)

5a + b = 16 → (5)

Subtract (4) from (5) term by term to eliminate b

2a = 6 ( divide both sides by 2 )

a = 3

Substitute a = 3 into (4) and evaluate for b

3(3) + b = 10

9 + b = 10 ( subtract 9 from both sides )

b = 1

Substitute a = 3, b = 1 into (1) and evaluate for c

3 + 1 + c = 1

4 + c = 1 ( subtract 4 from both sides )

c = - 3

Then a = 3, b = 1 and c = - 3

4 0

3 years ago

Solve the system of equations for the variables: 5x+2y=13 x+2y=9

r-ruslan [8.4K]

Answer:

x = 1

y = 4

Step-by-step explanation:

5x + 2y = 13

x + 2y = 9

Add both equations.

6x + 4y = 22

Solve for x.

6x = 22 - 4y

x = 22/6 - 4/6y

Put x as 22/6 - 4/6y in the second equation and solve for y.

22/6 - 4/6y + 2y = 9

-4/6y + 2y = 9 - 22/6

4/3y = 16/3

y = 16/3 × 3/4

y = 48/12

y = 4

Put y as 4 in the first equation and solve for x.

5x + 2(4) = 13

5x + 8 = 13

5x = 13 - 8

5x = 5

x = 5/5

x = 1

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3 years ago

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Alisha asked 120 students what kind of pet they liked the most. exactly 45% of the students said they liked dogs best. what was

Arada [10]

Answer is b. 54

Because 0.45 × 120 = 54

You're welcome, give thanks ;)

8 0

3 years ago

Help me to answer now ineed this <br> Please...

Vera_Pavlovna [14]

ANSWER TO QUESTION 1

$\frac{\frac{y^2-4}{x^2-9}} {\frac{y-2}{x+3}}$

Let us change middle bar to division sign.

$\frac{y^2-4}{x^2-9}\div \frac{y-2}{x+3}$

We now multiply with the reciprocal of the second fraction

$\frac{y^2-4}{x^2-9}\times \frac{x+3}{y-2}$

We factor the first fraction using difference of two squares.

$\frac{(y-2)(y+2)}{(x-3)(x+3)}\times \frac{x+3}{y-2}$

We cancel common factors.

$\frac{(y+2)}{(x-3)}\times \frac{1}{1}$

This simplifies to

$\frac{(y+2)}{(x-3)}$

ANSWER TO QUESTION 2

$\frac{1+\frac{1}{x}} {\frac{2}{x+3}-\frac{1}{x+2}}$

We change the middle bar to the division sign

$(1+\frac{1}{x}) \div (\frac{2}{x+3}-\frac{1}{x+2})$

We collect LCM to obtain

$(\frac{x+1}{x})\div \frac{2(x+2)-1(x+3)}{(x+3)(x+2)}$

We expand and simplify to obtain,

$(\frac{x+1}{x})\div \frac{2x+4-x-3}{(x+3)(x+2)}$

$(\frac{x+1}{x})\div \frac{x+1}{(x+3)(x+2)}$

We now multiply with the reciprocal,

$(\frac{(x+1)}{x})\times \frac{(x+2)(x+3)}{(x+1)}$

We cancel out common factors to obtain;

$(\frac{1}{x})\times \frac{(x+2)(x+3)}{1}$

This simplifies to;

$\frac{(x+2)(x+3)}{x}$

ANSWER TO QUESTION 3

$\frac{\frac{a-b}{a+b}} {\frac{a+b}{a-b}}$

We rewrite the above expression to obtain;

$\frac{a-b}{a+b}\div {\frac{a+b}{a-b}}$

We now multiply by the reciprocal,

$\frac{a-b}{a+b}\times {\frac{a-b}{a+b}}$

We multiply out to get,

$\frac{(a-b)^2}{(a+b)^2}$

ANSWER T0 QUESTION 4

To solve the equation,

$\frac{m}{m+1} +\frac{5}{m-1} =1$

We multiply through by the LCM of $(m+1)(m-1)$

$(m+1)(m-1) \times \frac{m}{m+1} + (m+1)(m-1) \times \frac{5}{m-1} =(m+1)(m-1) \times 1$

This gives us,

$(m-1) \times m + (m+1) \times 5}=(m+1)(m-1)$

$m^2-m+ 5m+5=m^2-1$

This simplifies to;

$4m-5=-1$

$4m=-1-5$

$4m=-6$

$\Rightarrow m=-\frac{6}{4}$

$\Rightarrow m=-\frac{3}{2}$

ANSWER TO QUESTION 5

$\frac{3}{5x}+ \frac{7}{2x}=1$

We multiply through with the LCM of $10x$

$10x \times \frac{3}{5x}+10x \times \frac{7}{2x}=10x \times1$

We simplify to get,

$2 \times 3+5 \times 7=10x$

$6+35=10x$

$41=10x$

$x=\frac{41}{10}$

$x=4\frac{1}{10}$

Method 1: Simplifying the expression as it is.

$\frac{\frac{3}{4}+\frac{1}{5}}{\frac{5}{8}+\frac{3}{10}}$

We find the LCM of the fractions in the numerator and those in the denominator separately.

$\frac{\frac{5\times 3+ 4\times 1}{20}}{\frac{(5\times 5+3\times 4)}{40}}$

We simplify further to get,

$\frac{\frac{15+ 4}{20}}{\frac{25+12}{40}}$

$\frac{\frac{19}{20}}{\frac{37}{40}}$

With this method numerator divides(cancels) numerator and denominator divides (cancels) denominator

$\frac{\frac{19}{1}}{\frac{37}{2}}$

Also, a denominator in the denominator multiplies a numerator in the numerator of the original fraction while a numerator in the denominator multiplies a denominator in the numerator of the original fraction.

That is;

$\frac{19\times 2}{1\times 37}$

This simplifies to

$\frac{38}{37}$

Method 2: Changing the middle bar to a normal division sign.

$(\frac{3}{4}+\frac{1}{5})\div (\frac{5}{8}+\frac{3}{10})$

We find the LCM of the fractions in the numerator and those in the denominator separately.

$(\frac{5\times 3+ 4\times 1}{20})\div (\frac{(5\times 5+3\times 4)}{40})$

We simplify further to get,

$(\frac{15+ 4}{20})\div (\frac{(25+12)}{40})$

$\frac{19}{20}\div \frac{(37)}{40}$

We now multiply by the reciprocal,

$\frac{19}{20}\times \frac{40}{37}$

$\frac{19}{1}\times \frac{2}{37}$

$\frac{38}{37}$

5 0

3 years ago