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3 years ago

Write the expression for 5 minus q

Mathematics

1 answer:

levacccp [35]3 years ago

8 0

Answer:

5 - q

Step-by-step explanation:

the term minus indicates a sign of subtraction, which looks like this

Therefore, the expression is 5 - q

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What is the inverse of f(x)=3x+6 ?

scZoUnD [109]

Answer:

$\boxed{\sf \ \ f^{-1}(x)=\dfrac{x-6}{3} \ \ }$

Step-by-step explanation:

hello,

we can write

$fof^{-1}(x)=x \ and \ fof^{-1}(x)=f(f^{-1}(x))=3f^{-1}(x)+6 \ so\\3f^{-1}(x)+6=x \ \ subtract \ 6\\3f^{-1}(x)=x-6 \ \ divide \ \ by \ \ 3\\ f^{-1}(x)=\dfrac{x-6}{3}$

hope this helps

3 0

3 years ago

EXTREMELY URGENT GEOMETRY! 15 POINTS!<br><br> PLEASE HELP

zhenek [66]

Answer:

20.3 units

Step-by-step explanation:

AB/CB = BD/AB

9/CB = 4/9

CB = 9×9/4

CB = 81/4 = 20¼ units

20.25 (20.3 units correct to 1 dp)

7 0

3 years ago

2x minus y equals 6, x plus y equals negative 3

kap26 [50]

Answer:

y=-4, x=1

Step-by-step explanation:

2x-y=6 x+y=-3

2(-3-y)-y=6 x=-3-y

-6-2y-y=6 x=-3+4

-3y=12 x=1

y=-4

8 0

3 years ago

Topic: The Quadratic Formula

Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

$x= \frac{-b +- \sqrt{b^2-4ac} }{2a}$

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² = x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

$x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\= -4 or 1$

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

$x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\$

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

$x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2$

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

7 0

3 years ago

A line, y = mx + b, passes through the point (10, 3) and is parallel to the line 2x+5y=12.

slava [35]

We have been given that a line $y=mx+b$ passes through the point (10, 3) and is parallel to the line $2x+5y=12$ . We are asked to find the y-intercept of the line.