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3 years ago

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A recipe includes 6 cups of flour and three fourths cup of butter butter. Write the ratio of the amount of flour to the amount o

f butter butter as a fraction in simplest form.

1 answer:

vodka [1.7K]3 years ago

4 0

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2=8−6/4x help please yall whats the x

Alenkasestr [34]

Answer:

x= 4

Step-by-step explanation:

How to solve your problem

Topics: Algebra

2=8−1⋅64⋅

2=8-1 \cdot \frac{6}{4} \cdot x

2=8−1⋅46⋅x

Solve

Divide the numbers

2=8−1⋅64⋅

2=8-1 \cdot {\color{#c92786}{\frac{6}{4}}} \cdot x

2=8−1⋅46⋅x

2=8−1⋅32

2=8-1 \cdot {\color{#c92786}{\frac{3}{2}}}x

2=8−1⋅23x

Multiply the numbers

2=8−1⋅32

2=8{\color{#c92786}{-1}} \cdot {\color{#c92786}{\frac{3}{2}}}x

2=8−1⋅23x

2=8−32

2=8{\color{#c92786}{-\frac{3}{2}}}x

2=8−23x

Combine multiplied terms into a single fraction

2=8−32

2=8-\frac{3}{2}x

2=8−23x

2=8+−3 2

2=8+\frac{-3x}{2}

2=8+2−3x

Find common denominator

2=8+−3 2

2=8+\frac{-3x}{2}

2=8+2−3x 2 =2⋅82+−3 2

2=\frac{2 \cdot 8}{2}+\frac{-3x}{2}

2=22⋅8+2−3x

Combine fractions with common denominator

2=2⋅82+−3 2

2=\frac{2 \cdot 8}{2}+\frac{-3x}{2}

2=22⋅8+2−3x

2=2⋅8−3 2

2=\frac{2 \cdot 8-3x}{2}

2=22⋅8−3x

Multiply the numbers

2=2⋅8-3 2

2=\frac{{\color{#c92786}{2}} \cdot {\color{#c92786}{8}}-3x}{2}

2=22⋅8−3x 2=16−3 2

2=\frac{{\color{#c92786}{16}}-3x}{2}

2=216−3x

Rearrange terms

2=16−3 2

2=\frac{{\color{#c92786}{16-3x}}}{2}

2=2 16−3x 2=−3+16 2

2=\frac{{\color{#c92786}{-3x+16}}}{2}

2=2−3x+16

Multiply all terms by the same value to eliminate fraction denominators

2=−3+16 2 2=\frac{-3x+16}{2}

2=2−3x+16

2⋅2=2(−3+16 2)

2 \cdot 2=2(\frac{-3x+16}{2})

2⋅2=2(2−3x+16)

Cancel multiplied terms that are in the denominator

2⋅2=2(−3+16 2)

2 \cdot 2=2(\frac{-3x+16}{2})

2⋅2=2(2−3x+16)

2⋅2=−3+16

2 \cdot 2=-3x+16

2⋅2=−3x+16

Multiply the numbers

2⋅2=−3+16

{\color{#c92786}{2}} \cdot {\color{#c92786}{2}}=-3x+16

2⋅2=−3x+16

4=−3+16

{\color{#c92786}{4}}=-3x+16

4=−3x+16

Subtract

16

from both sides of the equation

4=−3x+16

4−16=−3+16−16

4{\color{#c92786}{-16}}=-3x+16{\color{#c92786}{-16}}

4−16=−3x+16−16

Simplify

Subtract the numbers

−12=−3

Divide both sides of the equation by the same term

−12=−3

\frac{-12}{{\color{#c92786}{-3}}}=\frac{-3x}{{\color{#c92786}{-3}}}

−3−12=−3−3x

Simplify

Divide the numbers

Cancel terms that are in both the numerator and denominator

Move the variable to the left

= 4

8 0

3 years ago

True or False: The variables in the equation 4x-(5y)2=64x-(5y)2=6 are 4, 5, and 6

aleksley [76]

Answer: False

Step-by-step explanation:

4x - (5y) 2 = 64x - (5y) 2 = 6

Identify the variables:

4x, 5y, 64x

So, it would actually be:

4, 5, and 64

6 0

3 years ago

3 times negative 13/8

professor190 [17]

3•-13/8 is -4.875 or or -4 7/8

8 0

2 years ago

Read 2 more answers

Thea has a key on her calculator marked $\textcolor{blue}{\bf\circledast}$. If an integer is displayed, pressing the $\textcolor

worty [1.4K]

Answer:

$9$

Step-by-step explanation:

Given: Thea enters a positive integer into her calculator, then squares it, then presses the $\textcolor{blue}{\bf\circledast}$ key, then squares the result, then presses the $\textcolor{blue}{\bf\circledast}$ key again such that the calculator displays final number as $243$.

To find: number that Thea originally entered

Solution:

The final number is $243$.

As previously the $\textcolor{blue}{\bf\circledast}$ key was pressed,

the number before $243$ must be $324$.

As previously the number was squared, so the number before $324$ must be $18$.

As previously the $\textcolor{blue}{\bf\circledast}$ key was pressed,

the number before $18$ must be $81$

As previously the number was squared, so the number before $81$ must be $9$.

3 0

3 years ago

Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$

5 0

1 year ago