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Ksivusya [100]
3 years ago
14

Part

Mathematics
1 answer:
lesantik [10]3 years ago
3 0
<span>y = 2−x and y = 4x + 3

Let I be the point of intersection:

I belongs to the line y=2-x and at the same time I belongs to y=4x+3, then the coordinates of I are the same for y =2-x & y= 4x+2, in short if we replace the coordinates in y = 2-x & in y=4x+3 by their respective values, we will find an equality.

b) <span>2−x = 4x + 3

Replace x  in  the equation 2−x = 4x + 3  with the here below values to find if an equality exists

for x = -3 then </span> 2-(-3) = 4(-3)+3 → 5 = -9 IMPOSSIBE, it's not an equality

for x = 2 then  2-2 = 4(2)+3 →<span> 0 = 11 IMPOSSIBE, it's not an equality

</span>for x = 1 then  2-(1) = 4(1)+3 →<span> 1 = 7 IMPOSSIBE, it's not an equality

</span>for x = - 1 then  2-(-1) = 4(-1)+3 →<span> 3 = -1 IMPOSSIBE, it's not an equality

and you can replace x with all integers from - 3 to + 3 and you will find an INEQUALITY, so all these values are not a solution of thev equation

c. Solving the equation </span><span>2−x = 4x + 3 
</span><span>2−x = 4x + 3  

1st add x to both sides: 2-x + x = 4x + 3 + x </span>→2 = 5x + 3

2nd to this new equation <span>2 = 5x + 3, subtract 3 from both sides:

</span><span>2 - 3= 5x + 3 - 3 </span>→-1 = 5x

3rd in this new equation -1=5x, divide both sides by 5 → -1/5 = x
 And x = -1/5 is the solution of the system. To find the y value, you replace in any of the 2 equation  <span>y = 2−x and y = 4x + 3, x by its value (-1/5) and you will find y = 11/5</span>




</span>
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Let (8,−3) be a point on the terminal side of θ. Find the exact values of cosθ, cscθ, and tanθ.
denis23 [38]

Answer:

\text{Cos}\theta=\frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{x}{R}=\frac{8}{\sqrt{73}}

\text{Csc}\theta=-\frac{\sqrt{73}}{3}

\text{tan}\theta =\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{y}{x}=\frac{-3}{8}

Step-by-step explanation:

From the picture attached,

(8, -3) is a point on the terminal side of angle θ.

Therefore, distance 'R' from the origin will be,

R = \sqrt{x^{2}+y^{2}}

R = \sqrt{8^{2}+(-3)^2}

  = \sqrt{64+9}

  = \sqrt{73}

Therefore, Cosθ = \frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{x}{R}=\frac{8}{\sqrt{73}}

Sinθ = \frac{\text{Opposite side}}{\text{Hypotenuse}}=\frac{y}{R}=\frac{-3}{\sqrt{73} }

tanθ = \frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{y}{x}=\frac{-3}{8}

Cscθ = \frac{1}{\text{Sin}\theta}=\frac{R}{y}=-\frac{\sqrt{73}}{3}

6 0
4 years ago
A solid object in the shape of a cylinder is shown below.
lozanna [386]

Answer:

0.08\ kg/m^3

Step-by-step explanation:

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#We calculate the volume of the cylinder:

V=\pi r^2 h, r=D/2=6/2=3\ cm\\\\=3.14\times 3^2\times 7\\\\=197.82\ m^3

Given the mass= 15.3kg and volume as above, we calculate density as:

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Oksanka [162]
When you multiply the equation out and you simplify like terms you get.


2x^2 + 6xy + 4y^2
3 0
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Oliga [24]

Answer:

y-5=1/4(x-3)

Step-by-step explanation:

first you need to find the slope.

5-7 over 3-6 = -1/-4 or 1/4

next put into point-slope form.

y-y1=m(x-x1)

or y-5=1/4(x-3)

7 0
4 years ago
7 points!!!!
AnnZ [28]

Answer:

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Step-by-step explanation:

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