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creativ13 [48]
3 years ago
12

If A = 2.54 and 20B=A what is the value of B?

Mathematics
1 answer:
mixas84 [53]3 years ago
8 0
0.127, 2.54/20 =20*0.127
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Find g(a - 1) when g(x) = 5x - 4.
pishuonlain [190]

Answer:

g(a-1) = 5a-9

Step-by-step explanation:

Hi there!

g(x) = 5x - 4

Replace all x's with a-1:

g(a-1) = 5(a-1) - 4\\g(a-1) = 5a-5 - 4\\g(a-1) = 5a-9

I hope this helps!

7 0
3 years ago
Can someone help me.
Helen [10]

The solutions of the equations for x are given as follows:

a. x = 36.

b. x = 15.

<h3>What is the solution for item a?</h3>

The equation for item a is given by:

x/3 - 5 = 7

Isolating x, we have that:

x/3 = 12

x = 3 x 12

x = 36.

<h3>What is the solution for item b?</h3>

The equation for item b is given by:

x/2 - 5 = 2.5

Isolating x, we have that:

x/2 = 7.5

x = 7.5 x 2

x = 15.

Hence the solutions are:

a. x = 36.

b. x = 15.

More can be learned about equations at brainly.com/question/25537936

#SPJ1

3 0
1 year ago
Use synthetic division to evaluate f(x) = –2x^3+ 3x – 6 when x = -5.
fgiga [73]

Answer:

-131

Step-by-step explanation:

Use -5 in place of x

f(x) = –2x^3+ 3x – 6

f(x) = -2*-5^3+3*5-6

6 0
2 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
Pls answer asap, show work pls :(
coldgirl [10]

Answer:

2.Solve for the first variable in one of the equations, then substitute the result into the other equation.

Point Form: ( 3 , 5 )

Equation Form: x = 3 , y = 5

3.Solve for the first variable in one of the equations, then substitute the result into the other equation.

Point Form: ( 2 , 8 )

4.Solve for the first variable in one of the equations, then substitute the result into the other equation.

Point Form: ( 2 , − 3 )

Step-by-step explanation:

6 0
3 years ago
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