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Vlada [557]
3 years ago
14

How do i solve using the rational root theorem. 2x3-9x2-11x+8=0

Mathematics
1 answer:
Lapatulllka [165]3 years ago
4 0
2(3)-9(2)-11x+8=0
(2)(3)-(9)(2)-11x+8=0
6+-18+-11x+8=0
(-11x)+(6+-18+8)=0 Combine Like Terms.
-11x+-4=0
-11x-4=0
Add 4 to both sides.
-11x-4+4=0+4
-11x=4
Divide both sides by -11.
-11x/-11=4/-11
x=-4/11

Hope this helps!

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The area of a parallelogram is 120 cm ^ 2 and the height is 30 cm . Find the corresponding base
const2013 [10]

Answer:

4 cm

Step-by-step explanation:

The Area of a parallelogram is given by :

Area = base * height

Area = 120 cm² ; height = 30

Area = base * height

120 = base * 30

Base = 120 / 30

Base of parallelogram = 4 cm

3 0
2 years ago
Suppose h(t) = -4t^2+ 11t + 3 is the height of a diver above the water (in
Bingel [31]
The diver will hit the water when the height is zero, so we will want to set our height function equal to zero
0 = -4t^2 + 11t + 3
we will then factor to find out value of t that make the function equal to zero
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4 0
2 years ago
Read 2 more answers
Find the diagonal of a rectangular frame which measures 77 in. by 36 in.
Cerrena [4.2K]

Answer:

85

Step-by-step explanation:

think about the diagonal of the frame, breaking up a rectangle into 2 right angled triangles. we are already given the legs of the triangle (77, and 36) and now we have to solve fro the hypotenuse (or the diagonal). so use pythagorean theorem :

77^2 + 36^2 = c^2

7225 = c^2

c = 85

5 0
3 years ago
0.008 is 1 tenth of what decimal
shepuryov [24]
0.08, Hope it helped
3 0
2 years ago
Please help me <br> Show your work <br> 10 points
Svet_ta [14]
<h2>Answer</h2>

After the dilation \frac{5}{3} around the center of dilation (2, -2), our triangle will have coordinates:

R'=(2,3)

S'=(2,-2)

T'=(-3,-2)

<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

(x,y)→(x-2, y+2)

Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor \frac{5}{3}. Therefore our second partial rule will be:

(x,y)→\frac{5}{3} (x-2,y+2)

(x,y)→(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

(x,y)→(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)

(x,y)→(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

Now that we have our rule, we just need to apply it to each point of our triangle to perform the required dilation:

R=(2,1)

R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})

R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})

R'=(2,3)

S=(2,-2)

S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

S'=(2,-2)

T=(-1,-2)

T'=(\frac{5}{3} (-1)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

T'=(-3,-2)

Now we can finally draw our triangle:

8 0
3 years ago
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