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3 years ago

Hanna baked a cake, she used 6 tenths of flour, 3 eggs and4 fifths of sugar. how man ingredients did she used..

Mathematics

2 answers:

matrenka [14]3 years ago

7 0

You need to add them up together so it will be 4 and 2\5

lidiya [134]3 years ago

5 0

Since the question wants to know how many ingredients she used, All we have to do is add them.

So as the question says, she used 6/10 flour, 3 eggs, and 4/5 sugar.

Now we have two fractions in this scenario. Let's make them have the same denominator. Either we divide 6/10 by 2 or multiply 4/5 by 2. Let's multiply.

4/5 × 2 = 8/10

Now we have 6/10, 3, and 8/10

Lets add them together.

3 + 6/10 = 3 and 6/10

Now 3 and 6/10 + 8/10 = 4 and 4/10

Simplify even further...

and your answer will be 4 and 2/5

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Betty has 32 ribbons. She gave .25 to sherry and gave .5 to Marcy. She kept the remaining. How many ribbons did each girl have?

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4 years ago

Simplify:<br> ly - 2x - 3y + 4x

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3 years ago

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Describe the steps required to determine the equation of a quadratic function given its zeros and a point.

faltersainse [42]

Answer:

Procedure:

1) Form a system of 3 linear equations based on the two zeroes and a point.

2) Solve the resulting system by analytical methods.

3) Substitute all coefficients.

Step-by-step explanation:

A quadratic function is a polynomial of the form:

$y = a\cdot x^{2}+b\cdot x + c$ (1)

Where:

$x$ - Independent variable.

$y$ - Dependent variable.

$a$ , $b$ , $c$ - Coefficients.

A value of $x$ is a zero of the quadratic function if and only if $y = 0$ . By Fundamental Theorem of Algebra, quadratic functions with real coefficients may have two real solutions. We know the following three points: $A(x,y) = (r_{1}, 0)$ , $B(x,y) = (r_{2},0)$ and $C(x,y) = (x,y)$

Based on such information, we form the following system of linear equations:

$a\cdot r_{1}^{2}+b\cdot r_{1} + c = 0$ (2)

$a\cdot r_{2}^{2}+b\cdot r_{2} + c = 0$ (3)

$a\cdot x^{2} + b\cdot x + c = y$ (4)

There are several forms of solving the system of equations. We decide to solve for all coefficients by determinants:

$a = \frac{\left|\begin{array}{ccc}0&r_{1}&1\\0&r_{2}&1\\y&x&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$a = \frac{y\cdot r_{1}-y\cdot r_{2}}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x+x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$a = \frac{y\cdot (r_{1}-r_{2})}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$b = \frac{\left|\begin{array}{ccc}r_{1}^{2}&0&1\\r_{2}^{2}&0&1\\x^{2}&y&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$b = \frac{(r_{2}^{2}-r_{1}^{2})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$c = \frac{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&0\\r_{2}^{2}&r_{2}&0\\x^{2}&x&y\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$c = \frac{(r_{1}^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x + x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

And finally we obtain the equation of the quadratic function given two zeroes and a point.

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3 years ago

Which of the following units for mass is an SI unit?

s2008m [1.1K]

Kilogram is an SI unit for mass.
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5 0

3 years ago

Read 2 more answers

Got a calc 1 question,

belka [17]

$v=\dfrac h{3x}\implies\dfrac{\mathrm dv}{\mathrm dt}=\dfrac{3x\frac{\mathrm dh}{\mathrm dt}-3h\frac{\mathrm dx}{\mathrm dt}}{3x^2}$

By the chain rule,

$\dfrac{\mathrm dh}{\mathrm dt}=\dfrac{\mathrm dh}{\mathrm dx}\dfrac{\mathrm dx}{\mathrm dt}$

We have

$h(x)=4e^{2x-6}-x^2+5\implies\dfrac{\mathrm dh}{\mathrm dx}=8e^{2x-6}-2x$

and we're given that $x$ changes at a constant rate of $\frac{\mathrm dx}{\mathrm dt}=0.2$ thousand people per minute, which means

$\dfrac{\mathrm dv}{\mathrm dt}=\dfrac{3x\left(8e^{2x-6}-2x\right)\left(0.2\frac{\text{thousand people}}{\rm min}\right)-3h\left(0.2\frac{\text{thousand people}}{\rm min}\right)}{3x^2}$

At the moment $x=4$ thousand people are in the park, we have $h(4)=4e^2-11$ , so we find

$\dfrac{\mathrm dv}{\mathrm dt}=\dfrac{12\left(8e^2-8\right)-3(4e^2-11)}{240}\dfrac{\text{thousand people}}{\rm min}=\dfrac{7(4e^2-3)}{80}\dfrac{\text{thousand people}}{\rm min}$

or approximately 2.324 thousand people per minute.

5 0

3 years ago