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3 years ago

Solve the equation: 7(3x+2)=18-(x+48)

Mathematics

2 answers:

Reil [10]3 years ago

7 0

X=22.2

do the math (I just added this so its valid)

uranmaximum [27]3 years ago

5 0

7(3x + 2) = 18 - (x + 48)

Distribute on both sides. On the left side, multiply each term in parentheses by 7. On the right side, a negative sign to the left of parentheses is as if a -1 is there, and you must multiply everything inside the parentheses by -1.

7(3x + 2) = 18 -1(x + 48)

21x + 14 = 18 - x - 48

21x + 14 = -x - 30

22x = -44

x = (-44)/22

x = -2

Answer: x = -2

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dedylja [7]

Answer:

Step-by-step explanation:

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3 years ago

2 1/6* 2 5/6 IN SIMPLEST FORM

andrezito [222]

Answer:

6 5/36

Step-by-step explanation:

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3 years ago

Find an equation in slope intercept form of the line that has a slope of -8 and passes through point A (1,5)

ollegr [7]

Answer:

8x+y-13=0

Step-by-step explanation:

let the given point be

A(1,5)=(x1,y1)

slope(m)= -8

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(y-y1)=m(x-x1)

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3 0

3 years ago

Read 2 more answers

Solve for x. enter your answer in interval notation using grouping symbols. x^2+15x<-36

Oliga [24]

X²+15x+36<0

at first solve quadratic equation

D=b²-4ac= 225-4*1*36= 81

x=(-b+/-√D)/2a
x=(-15+/-√81)/2= (-15+/-9)/2
x1=(-15-9)/2=-12
x2=(-15+9)/2=-3

we can write x²+15x+36<0 as (x+12)(x+3)<0

(x+12)(x+3)<0 can be 2 cases, because for product to be negative one factor should be negative , and second factor should be positive
1 case) x+12<0, and x+3>0,
x<-12, and x>-3
(-∞, -12) and(-3,∞) gives empty set

or second case) x+12>0 and x+3<0
x>-12 and x<-3
(-12,∞) and (-∞,-3) they are crossing , so (-12, -3) is a solution of this inequality

3 0

3 years ago

Which statements about the function are true? Select two

iogann1982 [59]

Answer:

The vertex of the function is at (1,-25)

Step-by-step explanation:

I think your question missed key information, allow me to add in and hope it will fit the orginal one.

Part of the graph of the function f(x) = (x + 4)(x-6) is shown below.

Which statements about the function are true? Select two

options.

The vertex of the function is at (1,-25).

The vertex of the function is at (1,-24).

The graph is increasing only on the interval -4< x < 6.

The graph is positive only on one interval, where x <-4.

The graph is negative on the entire interval

My answer:

Given the factored form of the function:

f(x) = (x + 4)(x-6)

<=> f(x) = $x^{2} - 2x -24$

We will convert to vertex form

<=> f(x) = ( $x^{2} - 2x +1$ ) - 25

<=> f(x) = $(x-1)^{2} -25$

=> the vertex of the function is: (1,-25)

We choose: a. The vertex of the function is at (1,-25)

Let analyse other possible answers:

c. The graph is increasing only on the interval -4< x < 6.

Because the parameter a =1 so the graph open up all over its domain and the vertex is the lowest point.

So the graph is increasing in the domain (1, +∞)

=> C is wrong

d. The graph is positive only on one interval, where x <-4

Wrong, The graph is positive only on one interval, where x > 6

e. The graph is negative on the entire interval

Wrong, The graph is negative only on one interval, where -4< x < 6.

7 0

4 years ago