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Bumek [7]
3 years ago
13

Which number line represents the solutions to [x - 5] = 1?

Mathematics
1 answer:
Strike441 [17]3 years ago
7 0

Answer:

there should only be two points at 6 and 4

Step-by-step explanation:

[x-5]=1

x-5=+ 1 or -1

x-5= 1

x-5= -1

x=6 or 4

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(a 3 - 2a + 5) - (4a 3 - 5a 2 + a - 2)
Andreas93 [3]

Answer:

- 3a³ + 5a² - 3a + 7

Step-by-step explanation:

Given

(a³ - 2a + 5) - (4a³ - 5a² + a - 2)

Distribute both parenthesis noting the second is distributed by - 1

= a³ - 2a + 5 - 4a³ + 5a² - a + 2 ← collect like terms

= (a³ - 4a³ ) + 5a² + (- 2a - a) + (5 + 2)

= - 3a³ + 5a² - 3a + 7

3 0
3 years ago
Use the distributive property to write an equivalent expression for -1/2 (4 minus 6x)
Softa [21]

Answer:

<h2>3x - 2</h2>

Step-by-step explanation:

<em>The distributive property</em>:

<em>a(b + c) = ab + ac</em>

-\dfrac{1}{2}(4-6x)=\left(-\dfrac{1}{2}\right)(4)+\left(-\dfrac{1}{2}\right)(-6x)=-2+3x

3 0
3 years ago
15.09 ÷ 0.01 = ?<br> 267.4 ÷ 1/10 = ?
Masja [62]

Answer:

Dividing by a fraction is multiplying by the reciprocal

15.09/1/100=15.09*100=1509

267.4/1/10=267.4*10=2674

8 0
3 years ago
Read 2 more answers
Which pair of lines are perpendicular
Gnesinka [82]

The answer is A. A perpendicular equation will have the opposite and reciprocal slope as the original equation.


Once converted to the slope-intercept form of y = mx+b, we find the two equations for A are y = 2x-1 and y = (-1/2)x + 3/2. The opposite reciprocal of the first equation's slope (m in the equation form given) is -1/2, which as you can see if the slope of the other equation.

5 0
3 years ago
Read 2 more answers
Evaluate the given integral by changing to polar coordinates. sin(x2 + y2) dA R , where R is the region in the first quadrant be
Leona [35]

In polar coordinates, the region R is the set of points

\left\{(r,\theta)\mid3\le r\le5,\,0\le\theta\le\dfrac\pi2\right\}

and we have x^2+y^2=r^2 and \mathrm dA=r\,\mathrm dr\,\mathrm d\theta. So the integral, converted to polar, is

\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_3^5r\sin r^2\,\mathrm dr\,\mathrm d\theta=\frac\pi2\int_3^5r\sin r^2\,\mathrm dr

Substitute s=r^2 to get

=\displaystyle\frac\pi4\int_9^{25}\sin s\,\mathrm ds=\frac{(\cos9-\cos25)\pi}4

5 0
3 years ago
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