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Len [333]
2 years ago
7

Which graph shows the solution set for Negative 1.1 x + 6.4 greater-than negative 1.3? A number line going from negative 10 to 0

. An open circle is at negative 7. Everything to the left of the circle is shaded. A number line going from negative 10 to 0. An open circle is at negative 7. Everything to the right of the circle is shaded. A number line going from 0 to 10. An open circle is at 7. Everything to the left of the circle is shaded. A number line going from 0 to 10. An open circle is at 7. Everything to the right of the circle is shaded.
Mathematics
2 answers:
konstantin123 [22]2 years ago
8 0

Answer:

its c

Step-by-step explanation:

took exam

Rufina [12.5K]2 years ago
4 0

Answer:

A number line going from 0 to 10. An open circle is at 7. Everything to the left of the circle is shaded.

Step-by-step explanation:

We have the following inequality:

-1.1x + 6.4 > -1.3

to solve it, we need to isolate x, as follows:

-1.1x > -1.3 - 6.4

-1.1x > -7.7

x < -7.7/-1.1     (notice that while dividing with a negative number, the sign          changes)

x < 7

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Given directed line segment QS , find the coordinates of R
Degger [83]

Answer:

The answer is below

Step-by-step explanation:

The question is not complete, what are the coordinates of point Q and R. But I would show how to solve this.

The location of a point O(x, y) which divides line segment AB in the ratio a:b with point A at (x_1,y_1) and B(x_2,y_2) is given by the formula:

x=\frac{a}{a+b}(x_2-x_1)+x_1\\ \\y=\frac{a}{a+b}(y_2-y_1)+y_1

If point Q is at (x_1,y_1) and S at (x_2,y_2)  and R(x, y) divides QS in the ratio QR to RS is 3:5, The coordinates of R is:

x=\frac{3}{3+5}(x_2-x_1)+x_1=\frac{3}{8}(x_2-x_1)+x_1\\ \\y=\frac{3}{3+5}(y_2-y_1)+y_1=\frac{3}{8}(y_2-y_1)+y_1

Let us assume Q(−9,4) and S(7,−4)

x=\frac{3}{8}(7-(-9))+(-9)=\frac{3}{8}(16)-9=-3\\\\y=\frac{3}{8}(-4-4)+4=\frac{3}{8}(-8)+4=1

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Idk sorryy but that seems astronbal
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<span>The expression given (27x^2)z/(-3x^2)(z^6), can be simplify as below:

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<span>
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