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3 years ago

Which is the solution set of the compound inequality 3.5x-10>-3 and 8x-9<39?

Mathematics

2 answers:

Goshia [24]3 years ago

6 0

Solve the two inequalities.
3.5x-10>-3
3.5x>7
x>2

8x-9<39
8x<48
x<6

Hope it helped!

Andreyy893 years ago

5 0

Answer:

The solution set of the given compound inequality is x ∈ (2 , 6)

Step-by-step explanation:

The inequalities are given to be : 3.5x - 10 > -3 and 8x - 9 < 39

First take : 3.5x - 10 > -3

Adding 10 on both the sides

⇒ 3.5x - 10 + 10 > -3 + 10

⇒ 3.5x > 7

Now, dividing by 3.5 both the sides

⇒ x > 2

Now, 8x - 9 < 39

Adding 9 on both the sides

⇒ 8x - 9 + 9 < 39 + 9

⇒ 8x < 48

Now, dividing by 8 on both the sides

⇒ x < 6

Thus, The solution set of the given compound inequality is x ∈ (2 , 6)

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miskamm [114]

Answer:

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Step-by-step explanation:

I'm going to presume you want this linear equation in slope-intercept form, so start with:

$6=4x+9y$

Subtract $9y$ from both sides of the equation:

$-9y+6=4x$

Subtract $6$ from both sides of the equation:

$-9y=4x-6$

Divide both sides of the equation by the coefficient of $y$ , which is $-9$ :

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Simplify:

$y=-\frac{4}{9}x +\frac{2}{3}$

5 0

3 years ago

Read 2 more answers

Find an equation of the circle that satisfies the given conditions.

hodyreva [135]

The equation of a circle:
$(x-h)^2+(y-k)^2=r^2$
(h,k) - the coordinates of the centre
r - the radius

The midpoint of the diameter is the centre of a circle.
The coordinates of the midpoint:
$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$
(x₁,y₁), (x₂,y₂) - the coordinates of endpoints

$P(-1,1) \\
x_1=-1 \\ y_1=1 \\ \\ Q(5,9) \\ x_2=5 \\ y_2=9 \\ \\
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The centre of the circle is (2,5).

The radius is the distance between an endpoint of the diameter and the centre.
The formula for distance:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(-1,1) \\ x_1=-1 \\ y_1=1 \\ \\ (2,5) \\ x_2=2 \\ y_2=5 \\ \\ d=\sqrt{(2-(-1))^2+(5-1)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

The radius is 5.

$(x-2)^2+(y-5)^2=5^2 \\
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lisov135 [29]

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Step-by-step explanation:

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