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mina [271]
3 years ago
4

Kate begins solving the equation StartFraction 2 Over 3 EndFraction left-parenthesis 6 x minus 3 right-parenthesis equals StartF

raction one-half EndFraction left-parenthesis 6 x minus 4 left-parenthesis.(6x – 3) = StartFraction 2 Over 3 EndFraction left-parenthesis 6 x minus 3 right-parenthesis equals StartFraction one-half EndFraction left-parenthesis 6 x minus 4 left-parenthesis.(6x – 4). Her work is correct and is shown below. StartFraction 2 Over 3 EndFraction left-parenthesis 6 x minus 3 right-parenthesis equals StartFraction one-half EndFraction left-parenthesis 6 x minus 4 left-parenthesis.(6x – 3) = StartFraction 2 Over 3 EndFraction left-parenthesis 6 x minus 3 right-parenthesis equals StartFraction one-half EndFraction left-parenthesis 6 x minus 4 left-parenthesis.(6x – 4) 4x – 2 = 3x – 2 When she adds 2 to both sides, the equation 4x = 3x results. Which solution will best illustrate what happens to x ? The equation has infinite solutions. The equation has one solution: x = 0. The equation has one solution: x = StartFraction 4 Over 3 EndFraction.. The equation has no solution.
Mathematics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

(B)The equation has one solution: x = 0.

Step-by-step explanation:

The equation Kate is solving is: \dfrac{2}{3} (6x-3)=\dfrac{1}{2}(6x-4)

\dfrac{2(6x-3)}{3}=\dfrac{(6x-4)}{2}\\\dfrac{2*3(2x-1)}{3}=\dfrac{2(3x-2)}{2}\\4x-2=3x-2\\$Adding two both sides$\\4x=3x\\4x-3x=0\\x=0

Therefore, the equation has one solution: x = 0.

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Solve △PQR. Round function values to 4 decimal places. Round final answers to the nearest tenth.
mestny [16]

Answer:

The length of q is 7.6 units

The measure of angle R is 31.9°

The measure of angle P is 66.1°

Step-by-step explanation:

Let us use the sine and cosine rules to solve the triangle

In Δ PQR,

p is the opposite side to ∠P

q is the opposite side to ∠Q

r is the opposite side to ∠R

∵ p = 7 units

∵ r = 4 units

∵ m∠Q = 82°

- Angle Q is between p and r so let us find q using cosine rule

∵ q² = p² + r² - 2(p)(r) cos(∠Q)

- Substitute the values of p, r and ∠Q in the rule

∴ q² = (7)² + (4)² - 2(7)(4) cos(82°)

∴ q² = 49 + 16 - 56 cos(82°)

∴ q² = 57.2063

- Take √  for both sides

∴ q = 7.5635

- Round it to the nearest tenth

∴ The length of q is 7.6 units

Now let us use the sine rule to find the measures of ∠R and ∠P

∵ \frac{q}{sin(Q)}=\frac{r}{sin(R)}=\frac{p}{sin(P)}

- Let us find sin(R)

∴  \frac{7.5635}{sin(82)}=\frac{4}{sin(R)}

- By using cross multiplication

∴ 7.5635 × sin(R) = 4 × sin(82)

∴ 7.5635 sin(R) = 4 sin(82)

- Divide both sides 7.5635

∴ sin(R) = 0.5285

- Use sin^{-1} to find m∠R

∵ m∠R = sin^{-1} (0.5285)

∴ m∠R = 31.9042

- Round it to the nearest tenth

∴ The measure of angle R is 31.9°

∵ The sum of the measures of the interior angles of a Δ is 180°

∴ m∠P + m∠Q + m∠R = 180°

∴ m∠P + 82 + 31.9 = 180

∴ m∠P + 113.9 = 180

- Subtract 113.9 from both sides

∴ m∠P = 66.1°

∴ The measure of angle P is 66.1°

4 0
3 years ago
Please help I don’t know the answer<br> The last option is none of the above
saw5 [17]
The first choice because of rise over run and y-intercepts
5 0
3 years ago
Read 2 more answers
What is 20/36 reduced
likoan [24]
20/36 reduced is 5/9. hope this helped!  :)
7 0
3 years ago
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PLEASE help me with the answers!!!
jasenka [17]
100=1,478,300
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5 0
4 years ago
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Write the standard equation for the circle center (-6,8) that passes through (0,0)
Reil [10]

Answer:

D

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

here (h, k) = (- 6, - 8), thus

(x + 6)² + (y + 8)² = r²

The radius is the distance from the centre to a point on the circle

Calculate r using the distance formula

r = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 6, - 8) and (x₂, y₂ ) = (0, 0)

r = \sqrt{(0+6)^2+(0+8)^2} = \sqrt{36+64} = \sqrt{100} = 10

Hence

(x + 6)² + (y + 8)² = 100 → D

4 0
3 years ago
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