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adell [148]
3 years ago
7

5/3(6x+3)<2x-7 what is the solution for the iniquality

Mathematics
1 answer:
leva [86]3 years ago
8 0
Isolate "x" onto one side:

5/3 * (6x+3) < 2x - 7

5(6x+3) < 3(2x-7)

30x + 15 < 6x - 21

24x < -36

x < -1.5
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Ratio of 4 dollars to 6 dollars.
NNADVOKAT [17]
2:3 ration when you simplify it out.
4 0
3 years ago
Read 2 more answers
How many feet long is the fence ?
grandymaker [24]

Answer:

50 feet long

Step-by-step explanation:

Use the Pythagorean theorem to find the value of the missing side. The pythagoreon theorem only works with right triangles such as this one.

a²+b²=c²

Where a and b are length of legs (sides) of a triangle. And c is the length of the hypotenuse. The hypotenuse is the side across from the right angle.

We know by looking at the image posted in the question, that one leg of the triangle is 14 feet. And another leg is 48 feet. (Length of picnic area) (length of campground)

14²+48²=c²

Simplify

196+2304=2500

Find square root of 2500

√(2500)=50

So 14²+48²=50²

The missing side is 50 feet long. The fence is 50 feet long.

Attached is another example.

7 0
3 years ago
What is the slope of (-5,3) and (7,9)
Softa [21]

Answer:

-35

___

27

Step-by-step explanation:

Start by setting up your pairs (x,y)

-5 & 7 are x

3 & 9 are y

X * X

---------- ÷

Y * Y

(-5, 7)

---------

(3, 9)

-35

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27

3 0
3 years ago
Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -
GaryK [48]

Consider all parabolas:

1.

y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

2.

y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

3.

y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

4.

y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

5.

y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

6.

y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0
3 years ago
In your own words, write a verbal expression for the following algebraic expression.<br> 18p
maxonik [38]

Step-by-step explanation:

18p is a multiplication expression in itself.

It could be written as 18 x p. I am confused on what you mean as 'verbal' so i'm gonna take a wild guess and say word form.

eighteen times p???

Sorry if it's wrong.

6 0
3 years ago
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