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2 years ago

A recipe calls for 3.75 cups of flower. What is this amount as a mixed number in simplest form

Mathematics

1 answer:

Nadusha1986 [10]2 years ago

8 0

3.75 as a mixed number is 3 3/4

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What is MEAN ABSOLUTE DEVIATION and how do you find it?

Fynjy0 [20]

Step-by-step explanation:

Step 1: Calculate the mean. Step 2: Calculate how far away each data point is from the mean using positive distances. These are called absolute deviations. Step 3: Add those deviations together. Step 4: Divide the sum by the number of data points.

8 0

2 years ago

Gayle installed a rectangular section of hardwood flooring measuring 12 ft by 12 ft in her family room. She plans on increasing

icang [17]

The ansa is 256=(12+x)(12+x)

6 0

2 years ago

vanessa invested $2500 into an account that will increase in value 3.5% each year. write an exponential function, then find the

riadik2000 [5.3K]

Answer: A = 2500 (1.035)^n

A= $ 4,974.47

Step-by-step explanation:

A=P(1+r)^n

A= Amount

P= Principal

R= rate

N= # of years

A= 2500(1.035)^n

N=20

2500(1.035)^20

A= 4,974.47

4 0

3 years ago

If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c

dolphi86 [110]

Given:

$(ax+2)(bx+7)=15x^2+cx+14$

And

$a+b=8$

Required:

To find the two possible values of c.

Explanation:

Consider

$\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}$

$\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}$

And also given

$a+b=8---(2)$

Now from (1) and (2), we get

$\begin{gathered} a+\frac{15}{a}=8 \\ \\ a^2+15=8a \\ \\ a^2-8a+15=0 \end{gathered}$ $a=3,5$

Now put a in (1) we get

$\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}$

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

$\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}$

When a=5 and b=3, we have

$\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}$

Final Answer:

The option D is correct.

31 and 41

8 0

1 year ago

Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, wh

fredd [130]

With

$\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k$

we have

$\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

The vector field evaluated over this parameterization is

$\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k$

so the line integral is

$\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

$=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4$

6 0

3 years ago