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1 year ago

Solve the following equation for a. -4G+1/8Qa=7 Answer: a=?

Mathematics

1 answer:

kolbaska11 [484]1 year ago

6 0

By making a the subject of formula in the given equation, the value "a" is equal to 1/(56Q + 32GQ).

<h3>What is an equation?</h3>

An equation can be defined as a mathematical expression which shows that two (2) or more thing are equal.

In this exercise, you're required to solve for a from the given mathematical expression (equation) by making it the subject of formula. This ultimately implies that, all the other variables would be defined in terms of a and they would all be on the same side of the "equal to" symbol.

Making a the subject of formula, we have:

-4G + 1/8Qa = 7

1/8Qa = 7 + 4G

Multiplying both sides by 8Q, we have:

8Q × 1/8Qa = (7 + 4G) × 8Q

1/a = 56Q + 32GQ

a = 1/(56Q + 32GQ).

Read more on subject of formula here: brainly.com/question/21140562

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I need the answers to this please and thank you:)

Liono4ka [1.6K]

Q # 1

Explanation

Given the parabola

$f\left(x\right)=\left(x-3\right)^2-1$

Openness

It OPENS UP, as 'a=1' is positive.

Finding Vertex

The vertex of an up-down facing parabola of the form

$y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

$\mathrm{Rewrite}\:y=\left(x-3\right)^2-1\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=x^2-6x+8$

$a=1,\:b=-6,\:c=8$

$x_v=-\frac{\left(-6\right)}{2\cdot \:1}$

$x_v=3$

Finding $y_v$

$y_v=3^2-6\cdot \:3+8$

$y_v=-1$

So vertex is:

$\left(3,\:-1\right)$

Horizontal Translation

$y=\left(x-3\right)^2$ moves the graph RIGHT 3 units.

Vertical Translation

$f\left(x\right)=\left(x-3\right)^2-1$ moves the graph DOWN 1 unit.

Stretch or Compress Vertically

As $a = 1$ , so it does not affect the stretchiness or compression.

Q # 2

Explanation:

$f\left(x\right)=-\left(x+1\right)^2-2$

Openness

It OPENS DOWN, as 'a=-1' is negative.

Vertex

$\mathrm{Rewrite}\:y=-\left(x+1\right)^2-2\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=-x^2-2x-3$

$a=-1,\:b=-2,\:c=-3$

$x_v=-\frac{\left(-2\right)}{2\left(-1\right)}$

$x_v=-1$

$\mathrm{Plug\:in}\:\:x_v=-1\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=-2$

So vertex is:

$\left(-1,\:-2\right)$

Horizontal Translation

$y=\left(x+1\right)^2$ moves the graph LEFT 1 unit.

Vertical Translation

$f\left(x\right)=\left(x+1\right)^2-2$ moves the graph DOWN 2 unit.

Stretch or Compress Vertically

As $a = -1$ < 0, so it is either stretched or compressed.

Q # 3

Explanation:

$f\left(x\right)=\frac{1}{3}\left(x-4\right)^2+6$

It OPENS UP, as 'a=1/3' is positive.

Vertex

$\mathrm{Rewrite}\:y=\frac{1}{3}\left(x-4\right)^2+6\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=\frac{1\cdot \:x^2}{3}-\frac{8x}{3}+\frac{34}{3}$

$a=\frac{1}{3},\:b=-\frac{8}{3},\:c=\frac{34}{3}$

$x_v=-\frac{\left(-\frac{8}{3}\right)}{2\left(\frac{1}{3}\right)}$

$x_v=4$

Finding $y_v$

$y_v=\frac{1\cdot \:4^2}{3}-\frac{8\cdot \:4}{3}+\frac{34}{3}$

$y_v=6$

So vertex is:

$\left(4,\:6\right)$

Horizontal Translation

$f\left(x\right)=\left(x-4\right)^2$ moves the graph RIGHT 4 units.

Vertical Translation

$f\left(x\right)=}\left(x-4\right)^2+6$ moves the graph UP 6 unit.

Stretch or Compress Vertically

As $a=\frac{1}{3}$ , so it the graph is vertically compressed by a factor of 1/3.

Check the attached comparison graphs.

Q # 4

Explanation:

Given the function

$f\left(x\right)=-\left(x+3\right)^2$

It OPENS DOWN, as 'a=-1' is negative.

Vertex

The vertex of an up-down facing parabola of the form $y=a\left(x-m\right)\left(x-n\right)$

is the average of the zeros $x_v=\frac{m+n}{2}$

$y=-\left(x+3\right)^2$

$a=-1,\:m=-3,\:n=-3$

$x_v=\frac{m+n}{2}$

$x_v=\frac{\left(-3\right)+\left(-3\right)}{2}$

$x_v=-3$

Finding $y_v$

$y_v=-\left(-3+3\right)^2$

$y_v=0$

So vertex is:

$\left(-3,\:0\right)$

Horizontal Translation

$y=\left(x+3\right)^2$ moves the graph LEFT 3 units.

Vertical Translation

$y=\left(x+3\right)^2$ does not move the graph vertically.

Stretch or Compress Vertically

As $a=-1$ , so it the graph is either vertically stretched or compressed.

Q # 5

Explanation:

$f\left(x\right)=\left(x+5\right)^2-3$

Openness

It OPENS UP, as 'a=1' is positive.

Vertex

$\mathrm{Rewrite}\:y=\left(x+5\right)^2-3\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=x^2+10x+22$

$a=1,\:b=10,\:c=22$

$x_v=-\frac{10}{2\cdot \:1}$

$x_v=-5$

Finding $y_v$

$y_v=\left(-5\right)^2+10\left(-5\right)+22$

So vertex is:

$\left(-5,\:-3\right)$

Horizontal Translation

$f\left(x\right)=\left(x+5\right)^2$ moves the graph LEFT 5 units.

Vertical Translation

$f\left(x\right)=\left(x+5\right)^2-3$ moves the graph DOWN 3 unit.

Stretch or Compress Vertically

As $a = 1$ , so it does not affect the stretchiness or compression.

Check the attached comparison graphs.

Q # 6

THE DETAILS OF COMPLETE SOLUTION OF QUESTION 6 IS ATTACHED IN THE DIAGRAM AS THE 5000 CHARACTERS WERE ALREADY FILLED. SO, I solved via the attached figure.

SO, PLEASE CHECK THE LAST FIGURE TO FIND THE COMPLETE SOLUTION OF THE Q#6.

6 0

3 years ago

Dan sells beaded necklaces. Each large necklace sells for $5.30 and each small necklace sells $3.90. How much will he earn from

ycow [4]

5.30 for 1 large and 19.50 for 5 smalls

5 0

3 years ago

Read 2 more answers

Write an expression for the calculation add 34 and 6 and then multiply 3

Ymorist [56]

We need to write an expression for "add 34 and 6 and then multiply 3"

"Add" represents + operation .

"multiply" represents × operation.

We need to add 34 and 6.

So, we can write 34+6 and then we need multiply the sum by 3.

So, we need to keep above addition of 34 and 6 in parantehsis. We get

(34+6) × 3.

So, the final expression is (34+6) × 3.

3 0

3 years ago

[15 Points] Solve for x.<br> <img src="https://tex.z-dn.net/?f=45x%2B34%3D24x%2B16" id="TexFormula1" title="45x+34=24x+16" alt="

sergiy2304 [10]

Answer:

6/7

Step-by-step explanation:

45x+34=24x+16

collect like terms, we now have;

45x-24x=16-34

21x=-18

X=18/21

X=6/7

8 0

3 years ago

I don't know if this is right... please someone help mee

worty [1.4K]

For the first circle, let's use the pythagorean theorem

$\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17$

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.

now, let's check for second circle

$\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938$

well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.

let's check the third circle

$\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}$

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.

6 0

3 years ago