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lord [1]
3 years ago
14

Jacques needs 5/8 kg of flour to make a loaf of bread

Mathematics
1 answer:
Semenov [28]3 years ago
6 0

Answer:

9/40 kg

Step-by-step explanation:

Given:

He needs 5/8 kg but he first tips only 2/5 kg

additional flour needed,

= amount needed - amount already tipped

= 5/8 - 2/5

= 25/40 - 16/40

= (25 - 16) / 40

= 9/40 kg

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Use the given information to write the equation of the parabola.
m_a_m_a [10]

Answer:

x² = -2y

Step-by-step explanation:

The focus is p away from the vertex, and so is the directrix.

To find the equation of the parabola, we must first determine if the parabola is horizontal or vertical.

  • Horizontal parabola [Standard form]: (y – k)² = 4p(x – h)
  • Vertical parabola [Standard form]: (x – h)² = 4p(y – k)

If the parabola is vertical, the directrix, and focus will have the same x value but different y value compared to the vertex (h, k). You can also tell if the directrix in in the form y = k – p, and if the focus is in the form (h, k + p).

Likewise, if the parabola is horizontal, the directrix, and focus will have the same y value but different x value compared to the vertex (h,k) . You can also tell if the directrix is in the form x = h – p, and if the focus is in the form (h + p, k).

For this problem, given that the vertex is at the origin (0,0), and that the focus is at the point (0, -½).

You can tell that the x value is the same for the vertex, and focus so this must be a vertical parabola. Because this is a vertical parabola, we can use the form mentioned as earlier (x – h)² = 4p(y – k).

If h = 0, and k = 0, the p value must be the difference between the k of the vertex, and the k of the focus: -½ - 0 → -½.

Now we can just plug in our known information to derive the equation!

h = 0, k = 0, p = -½ → (x - h)² = 4p(y - k) →

(x - 0)² = 4(-½)(y - 0) → x² = -2(y - 0) →

x² = -2y.

Also p = 1/4a, if you are wondering.

So because this is a vertical parabola, x² = -2y is generally the same as y = -1/2x² in standard quadratic form. I just like to think of the horizontal parabola as an inverse quadratic because it is like reflecting over the line y = x.

8 0
3 years ago
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18
GalinKa [24]

Answer:

18

ZAOC and ZCOB are supplementary angles.

Find the

6 0
2 years ago
Which integer is 30 closest to
vesna_86 [32]

Answer:

7,4,15,6

Step-by-step explanation:

7,4,15,6

4 0
3 years ago
Read 2 more answers
Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^
Gemiola [76]

Answer:

\rm \displaystyle y' =   2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x}

Step-by-step explanation:

we would like to figure out the differential coefficient of e^{2x}(1+\ln(x))

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

\displaystyle y =  {e}^{2x}  \cdot (1 +   \ln(x) )

to do so distribute:

\displaystyle y =  {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x}

take derivative in both sides which yields:

\displaystyle y' =  \frac{d}{dx} ( {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x} )

by sum derivation rule we acquire:

\rm \displaystyle y' =  \frac{d}{dx}  {e}^{2x}  +  \frac{d}{dx}   \ln(x)  \cdot  {e}^{2x}

Part-A: differentiating $e^{2x}$

\displaystyle \frac{d}{dx}  {e}^{2x}

the rule of composite function derivation is given by:

\rm\displaystyle  \frac{d}{dx} f(g(x)) =  \frac{d}{dg} f(g(x)) \times  \frac{d}{dx} g(x)

so let g(x) [2x] be u and transform it:

\displaystyle \frac{d}{du}  {e}^{u}  \cdot \frac{d}{dx} 2x

differentiate:

\displaystyle   {e}^{u}  \cdot 2

substitute back:

\displaystyle    \boxed{2{e}^{2x}  }

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

\displaystyle  \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)

let

  • f(x) \implies   \ln(x)
  • g(x) \implies    {e}^{2x}

substitute

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =  \frac{d}{dx}( \ln(x) ) {e}^{2x}  +  \ln(x) \frac{d}{dx}  {e}^{2x}

differentiate:

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =   \boxed{\frac{1}{x} {e}^{2x}  +  2\ln(x)  {e}^{2x} }

Final part:

substitute what we got:

\rm \displaystyle y' =   \boxed{2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x} }

and we're done!

6 0
3 years ago
Suppose x=7 is a solution to the equation 4x-2(x+a)=8 how can you find the value of a that make the equation true
marta [7]

Answer:

a = 3

Step-by-step explanation:

Since x = 7 is a solution, substitute x = 7 into the equation and solve for a

4(7) - 2(7 + a) = 8 , that is

28 - 2(7 + a) = 8 ( subtract 28 from both sides )

- 2(7 + a) = - 20 ( divide both sides by - 2 )

7 + a = 10 ( subtract 7 from both sides )

a = 3

8 0
2 years ago
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