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Lyrx [107]
3 years ago
11

Need help solving this !

Mathematics
1 answer:
Zigmanuir [339]3 years ago
4 0

Answer:

Step-by-step explanation:

To start of u have to subtract 150 with 180 which is 180 (cuz its a straight line and E to get angle S

Now that we know that 180-150= 30

We add S and R then subtract by 180

so 80+ 30= 110

then we said subtract 180

180-110= 70

Now we know that angle Q is 7-

Then the bottom is RS & SE (NOT REALLY SURE)

It is a Remote Interior Angle

IF RIGHT PLZ GIVE BRAINLIEST  

THANK U  

HAVE A GREAT DAY AND BE SAFE :)

XD

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What is the slope of the line represented by the equation y = –1/2x+1/4?
max2010maxim [7]

Answer:

A. -½

Step-by-step explanation:

In the Slope-Intercept Formula, <em>y = mx + b</em><em>,</em><em> </em><em>m</em><em> </em>represents the <em>rate</em><em> </em><em>of</em><em> </em><em>change</em><em> </em>[<em>slope</em>], so in this case, the slope is -½.

I am joyous to assist you anytime.

7 0
2 years ago
What are the solutions of the quadratic equation (x + 3)(x +3) = 49? Ax = -2 and x = -16 Bx = 2 and x = -10 Cx = 4 and x = -10 D
Ivahew [28]

Answer:

4 and -10

Step-by-step explanation:

\displaystyle (x + 3)(x +3) = 49 \\ x^2+3x+3x+9=49 \\ x^2 +6x+9=49 \\ x^2 + 6x + 9 - 49 = 0 \\x^2+6x-40=0 \\\\ \Delta=b^2-4ac \\ \Delta=6^2-4 \cdot 1 \cdot (-40) \\ \Delta=36+160 \\ \Delta=196 \\ \\ X_{1,2}=\frac{-b \pm \sqrt{\Delta} }{2a}  \\ \\ X_1=\frac{-b+\sqrt{\Delta} }{2a} = \frac{-6+14}{2} = \frac{8}{2}=4 \\ \\ X_2=\frac{-b-\sqrt{\Delta} }{2a}  = \frac{-6-14}{2}=\frac{-20}{2} = -10

4 0
3 years ago
-x - 4y = 16 find value of y when x equals -8
pashok25 [27]
The answer is -2......
8 0
3 years ago
Read 2 more answers
38. Evaluate f (3x +4y)dx + (2x --3y)dy where C, a circle of radius two with center at the origin of the xy
lina2011 [118]

It looks like the integral is

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy

where <em>C</em> is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. Then

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on <em>C</em> nor in the region bounded by <em>C</em>, so

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy

where <em>D</em> is the interior of <em>C</em>, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: -2\times \pi\times2^2 = -8\pi.

3 0
2 years ago
Someone help me out here.
Vladimir [108]
Cremini:
$11.25 for 2/3 lb
2/3 lb = $11.25
1 lb = 11.25 ÷ 2/3 = $16.88 (nearest cent)

Chanterelle:
$7.99 for 1/2 lb
1/2 lb = $7.99
1 lb = $7.99 x 2 = $15.98

Answer:  Chanterelle costs less. It costs $0.90 less than Cremini.

4 0
3 years ago
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