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Nikitich [7]
3 years ago
7

1. Select all the expressions that are equivalent to 4 - x.

Mathematics
1 answer:
viktelen [127]3 years ago
4 0

Step-by-step explanation:

It is required to find the expressions that are equivalent to 4-x. We can also write it as :

Option (b) : (4-x) = 4+(-x)

Option (c) : (4-x) = -x+4

Hence, the correct options are (B) and (C).

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Acoordinate grid is shown below: a coordinate grid from negative 2 to 0 to positive 2 is drawn. there are three grid lines between a whole unit on the grid. part a: which point represents the origin? (2 points) part b: starting from the origin, explain how to plot the following three points accurately: (2, −2) (−2, 1.5) (−1, fraction 2 over 4)

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3 years ago
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Word form and standard form 2,364
stiks02 [169]

Answer:

Standard form: 2,364 is the standard form of 2,364

Word form: Two thousand, three hundred and sixty-four

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Solve for x, show all of your work: -2x – 6 &lt; 12
vova2212 [387]
Add 6 to both sides so you have -2x = 18. If you divide both sides by -2 you will get x = -9
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3 years ago
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Lines m and n are parallel. Which of the other 5 named angles have a measure of 110°?
Cloud [144]

Answer:

2 and angle that is left of 5

Step-by-step explanation:

Angle 2 is vertical from 110, and vertical angles are congruent. From this, we can figure out that 2 and 3 (they're supplementary) are 70. Angles 5 and 3 are Alternate Interior Angles, and therefore must be congruent. The one to the left of 5 is supplementary to it, and therefore must be 110. Your answers are angle 2 and the angle that is left of 5.

4 0
4 years ago
Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1
erik [133]

Answer:

\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8},  \dfrac{5}{14})}

Step-by-step explanation:

Given that:

y =  x^{2/3} at y = 0 , x = 1

Then:

Area = \int^{1}_{0} x^{2/3} \ dx

Area = \begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0

Area = \dfrac{3}{5}

Then:

\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx

\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx

\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx

\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0

\overline x = \dfrac{5}{3} \times \dfrac{3}{8}

\overline x = \dfrac{5}{8}

Similarly;

\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix}  \ dx

\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix}  \ dx

\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix}  \ dx

\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2}  (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0

\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14}  (x^{7/3} ) \end {bmatrix} ^1_0

\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )

\overline y = \dfrac{5}{14}

Thus; \mathbf{(\overline x , \overline y ) = (\dfrac{5}{8},  \dfrac{5}{14})}

4 0
3 years ago
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