MATHEMATICS HIGH SCHOOL

Given Q = 3a + 5ac solve for a

Answers

Answer 1
Answer:

Hi Mikey


q=3a+5ac

You need to change the equation's side

5ac+3a=q

Factor out a

a(5c+3)=q

Now we need to divide both sides by 5c+3 so we can find the value for a

a(5c+3)/(5c+3)=q/(5c+3)

a= q/(5c+3)


I hope that's help !

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Find the unit rate. Round your answer to the nearest hundredth.3500 calories for 6 servings of pie
What is the equation of a horizontal line that passes through the point at (-2,-3)????
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What is the area of this figure?

Answers

Answer:

35 in^2

Step-by-step explanation:

15+20=35

Which products result in a perfect square trinomial? Check all that apply.A=(–x + 9)(–x – 9)
B=(xy + x)(xy + x)
C=(2x – 3)(–3 + 2x)
D=(16 – x2)(x2 – 16)
E=(4y2 + 25)(25 + 4y2)

Answers

The correct answers are:

B=(xy + x)(xy + x) ; C=(2x – 3)(–3 + 2x) ; and E=(4y² + 25)(25 + 4y²)

Explanation:

In order to have a perfect square trinomial, we must multiply two binomials that are exactly the same.  For (xy+x)(xy+x), are multiplying two identical binomials.

For (2x-3)(-3+2x), we are multiplying two binomials that are the same but written in a different order.  The same is true of (4y² + 25)(25 + 4y²).

Answer:

A and E and C

Step-by-step explanation:

A perfect square trinomial can be written as the square of a binomial.

Which statement describes the behavior of the function f(x)=3x/4-x?a. The graph approaches –3 as x approaches infinity.
b. The graph approaches 0 as x approaches infinity.
c. The graph approaches 3 as x approaches infinity.
d. The graph approaches 4 as x approaches infinity.

Answers

Answer:

The  graph approaches –3 as x approaches infinity. Option a is correct.

Step-by-step explanation:

The given function is

f(x)=(3x)/(4-x)

We have to find value of function as x approaches infinity. Take limit both sides as x approaches to infinity.

\lim_(x\rightarrow \infty)f(x)=\lim_(x\rightarrow \infty)(3x)/(4-x)

Taking x common from the denominator.

\lim_(x\rightarrow \infty)f(x)=\lim_(x\rightarrow \infty)(3x)/(x((4)/(x)-1))

Cancel out common factor x.

\lim_(x\rightarrow \infty)f(x)=\lim_(x\rightarrow \infty)(3)/((4)/(x)-1)

Apply limits.

\lim_(x\rightarrow \infty)f(x)=(3)/((4)/(\infty)-1)

\lim_(x\rightarrow \infty)f(x)=(3)/(0-1)

\lim_(x\rightarrow \infty)f(x)=-3

Therefore the  graph approaches –3 as x approaches infinity.
a. The graph approaches –3 as x approaches infinity.

f(x) = (3x)/(4-x) +3 -3 = (3x+12-3x)/(4-x) -3 = (12)/(4-x) -3

The four hand of a clock moves from 12 to 5 o'clock through how many degrees does it move?

Answers

to find the answer you have to first find how many degrees are there in between each number in the clock. the clock is 360°degrees and there are 12 numbers marked in a clock., so you have to divide 360 by 12 to get the number of degrees in between each number,
360°÷12=30° so there are 30°degrees between each number. 

if a clock hand moves from 12 to 5, it pass 5 numbers, so to get your final answer you have to multiply 30° by 5=150°.

so your answer is 150°degrees.
hope you can understand what i said. :)

What is X divided by 2?

Answers

Algebraically that would be expressed as (x)/(2)

A design engineer is mapping out a new neighborhood with parallel streets. If one street passes through (6, 4) and (5, 2), what is the equation for a parallel street that passes through (−2, 6)?A. y = 1 half x + 5
B. y - 1 half x + 1
C. y = 2x + 10
D. y = 2x − 14

Answers

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the first street.

(\stackrel{x_1}{6}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{2}) ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{2}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{5}-\underset{x_1}{6}}} \implies \cfrac{ -2 }{ -1 } \implies 2

so we are really looking for the equation of a line whose slope is 2 and it passes through (-2 , 6)

(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\hspace{10em} \stackrel{slope}{m} ~=~ 2 \n\n\n \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\n \cline{1-1} \n y-y_1=m(x-x_1) \n\n \cline{1-1} \end{array}\implies y-\stackrel{y_1}{6}=\stackrel{m}{2}(x-\stackrel{x_1}{(-2)}) \implies y -6 = 2 ( x +2) \n\n\n y-6=2x+4\implies {\Large \begin{array}{llll} y=2x+10 \end{array}}
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