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3 years ago

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Making coconut cookies. The recipe calls for 420 g of coconut and 120 g of sugar. Only has 252 g of coconut. How much sugar must

be used to keep the same ratio of coconut to sugar?

1 answer:

geniusboy [140]3 years ago

4 0

Answer: 72 grams.

Step-by-step explanation:

Let be "s" the amount of sugar in grams that must be used for 252 grams of coconut to keep the same ratio of coconut to sugar.

Knowing that in the recipe for the coconut cookies, should be 420 grams of coconut and 120 grams of sugar, and you only have 252 grams of coconut, you can set up this proportion to find "s":

$\frac{420grams}{120grams}=\frac{252grams}{s}$

Now, you need to solve for "s":

$\frac{420grams}{120grams}=\frac{252grams}{s}\\\\s(\frac{420grams}{120grams})=252grams\\\\s=(252grams)(\frac{120grams}{420grams})\\\\s=72grams$

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Which graph represents the solution to the given system?<br> Y=-x+5<br> y= 1/4 x+10

jeka57 [31]

Answer:

See attachment

Step-by-step explanation:

The given system is :

$y = - x + 5$

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

Let us equate both equations and solve for x.

$- x + 5 = \frac{1}{4}x + 10$

$- 4x + 20 = x + 40$

$20 - 40 =x + 4x$

$- 20 = 5x$

$x = - 4$

$y = - - 4 + 5 = 9$

The the two lines of this system will meet at (-4,9)

The graph is shown in the attachment.

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3 years ago

A 1/17th scale model of a new hybrid car is tested in a wind tunnel at the same Reynolds number as that of the full-scale protot

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Answer:

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ is approximately 0.0002

Step-by-step explanation:

The given Reynolds number of the model = The Reynolds number of the prototype

The drag coefficient of the model, $c_{m}$ = The drag coefficient of the prototype, $c_{p}$

The medium of the test for the model, $\rho_m$ = The medium of the test for the prototype, $\rho_p$

The drag force is given as follows;

$F_D = C_D \times A \times \dfrac{\rho \cdot V^2}{2}$

We have;

$L_p = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2 \times L_m$

Therefore;

$\dfrac{L_p}{L_m} = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2$

$\dfrac{L_p}{L_m} =\dfrac{17}{1}$

$\therefore \dfrac{L_p}{L_m} = \dfrac{17}{1} =\dfrac{\rho _p}{\rho _p} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_p} \right)^2 = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{F_p}{F_m} = \dfrac{c_p \times A_p \times \dfrac{\rho_p \cdot V_p^2}{2}}{c_m \times A_m \times \dfrac{\rho_m \cdot V_m^2}{2}} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}$

$\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2$

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$\dfrac{F_m}{F_p} = \left( \left\dfrac{1}{17} \right)^3$ = (1/17)^3 ≈ 0.0002

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ ≈ 0.0002.

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2 years ago

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blondinia [14]

Answer:

(1, -2)

Step-by-step explanation:

midpoint x = (x₁ + x₂) / 2

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$(\frac{-1 + 3}{2} , \frac{2 + -(6)}{2} )$

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3 years ago

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alisha [4.7K]

Answer:

Those graphs do not intersect.

Estes gráficos no se intersecciónan

Step-by-step explanation:

The intersection points are x for which:

$f(x) = g(x)$

In this question:

$f(x) = x^{2} + 6x - 7$

$g(x) = 4x - 10$

So

$x^{2} + 6x - 7 = 4x - 10$

$x^{2} + 2x + 3 = 0$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$x^{2} + 2x + 3 = 0$

So $a = 1, b = 2, c = 3$

$\bigtriangleup = b^{2} - 4ac = 2^{2} - 4*1*3 = -8$

Sincce $\bigtriangleup$ is negative, there are no solutions, which means that those graphs do not intersect.

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3 years ago