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3 years ago

Which expression is equivalent to 6(14)? 6(10 + 4) 6(10 + 14) 6(1 + 40) 6(10 + 40)

Mathematics

1 answer:

krok68 [10]3 years ago

8 0

The first one because of you add 10 and 4 you get 14. And they both multiply 6 by 14.

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The dot plot shows the number of hours, to the nearest hour, that a sample of 5th- and 7th-grade students spend watching televis

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Answer:5th grade range-7

7th grade range-7

The ratio variation is -1

Step-by-step explanation:just got the question right

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3 years ago

Read 2 more answers

Let divf = 6(5 − x) and 0 ≤ a, b, c ≤ 12. (a) find the flux of f out of the rectangular solid 0 ≤ x ≤ a, 0 ≤ y ≤ b, and 0 ≤ z ≤

dusya [7]

Continuing from the setup in the question linked above (and using the same symbols/variables), we have

$\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}(\nabla\cdot f)\,\mathrm dV$
$=\displaystyle6\int_{z=0}^{z=c}\int_{y=0}^{y=b}\int_{x=0}^{x=a}(5-x)\,\mathrm dx\,\mathrm dy\,\mathrm dz$
$=\displaystyle6bc\int_0^a(5-x)\,\mathrm dx$
$=6bc\left(5a-\dfrac{a^2}2\right)=3abc(10-a)$

The next part of the question asks to maximize this result - our target function which we'll call $g(a,b,c)=3abc(10-a)$ - subject to $0\le a,b,c\le12$ .

We can see that $g$ is quadratic in $a$ , so let's complete the square.

$g(a,b,c)=-3bc(a^2-10a+25-25)=3bc(25-(a-5)^2)$

Since $b,c$ are non-negative, it stands to reason that the total product will be maximized if $a-5$ vanishes because $25-(a-5)^2$ is a parabola with its vertex (a maximum) at (5, 25). Setting $a=5$ , it's clear that the maximum of $g$ will then be attained when $b,c$ are largest, so the largest flux will be attained at $(a,b,c)=(5,12,12)$ , which gives a flux of 10,800.

7 0

3 years ago

The point-slope form of a line is y - y1 = m(x-x1), where m is the slope and (x1,

Vlada [557]

Answer

False

FalseThe correct point slope form passing through points (x1,y1) is (y-y1)=m (x-x1)(y−y1)=m(x−x1)

4 0

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

Olegator [25]

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;

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

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

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

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

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

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

$\dfrac{F_p}{F_m} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}= \left (\dfrac{17}{1} \right)^2 \times \left( \left\dfrac{17}{1} \right) = 17^3$

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

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

5 0

3 years ago

Which is the value of this expression when j=-2

Nuetrik [128]

Answer:

Step-by-step explanation:

(jk^-2/j^-1k^-3)^3

(-2)(-1)^-2

------------- ^3

(-2)^-1(-1)^-3

switch negative exponenst to other side

(-2)(-2)(-1)^3

__________

(-1)^2

4(-1)

-----

= -4

REMEMBER all to the 3rd power!

(-4)^3

-64

3 0

3 years ago