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3 years ago

X + 34 = 40 what does "x" equal? explain

Mathematics

2 answers:

butalik [34]3 years ago

8 0

Hey there!

6 + 34 = 40

so x = 6

Hope this helps
Have a great day (:

navik [9.2K]3 years ago

8 0

<span>X + 34 = 40
</span>subtract 34 on both sides
<span>X + 34 - 34 = 40 - 34
simplify
</span><span>X = 6</span>

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3 years ago

Let g(x) = f(x+3) for some function f(x). Is g(x) a function? From the previous problem, let the domain of f(x) be [1, 4). What

padilas [110]

The domain of g(x) is [4, 7)

<h3>How to determine the domain of g(x)?</h3>

The function is given as:

g(x) = f(x + 3)

Since f(x) is a function, then g(x) is also a function

The domain of f(x) is given as:

[1, 4)

The equivalent of these values in g(x) are

x = 1 + 3 = 4

x = 4 + 3 = 7

Hence, the domain of g(x) is [4, 7)

Read more about domain at

brainly.com/question/1770447

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1 year ago

Solve the quadratic equation (2x-3)^2 = 6(3-2x)

LenKa [72]

Answer:

$x=\frac{3\sqrt{2}-3 }{2}$

$x=-\frac{3\sqrt{2}-3 }{2}$

Step-by-step explanation:

Open the brackets first:

$4x^2+9=18-12x$

Simplify:

$4x^2+12x-9=0$

It is determined that this can't be factored, so rewrite the equation so you can complete the square.

$x^2+3x=\frac{9}{4}$

Take the square of 1.5 and add it to both sides:

$x^2 + 3x+ \frac{9}{4} =\frac{9}4} +\frac{9}{4}$

Factor:

$(x+\frac{3}{2})^2=\frac{9}{2}$

$x+\frac{3}{2} = \frac{3\sqrt{2} }{2 }$

or $x+\frac{3}{2} =-\frac{3\sqrt{2} }{2}$

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or $x=-\frac{3\sqrt{2}-3 }{2}$

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2 years ago

If l || m find the value of y

Darina [25.2K]

That means they are parallel so it’s 6

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2 years ago

Find the average value of f(x+y)=sin(x+y) over

Troyanec [42]

For each given region $D$ , the average value of $f$ over $D$ is the integral of $f$ over $D$ divided by the area of $D$ . In both cases, $D$ is a rectangle so the area is trivial.

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$\displaystyle\int_0^{5\pi/3}\int_0^{\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$\displaystyle=\int_0^{5\pi/3}\left(\cos y-\cos\left(y+\dfrac\pi3\right)\right)\,\mathrm dy$

$=\sin\dfrac{5\pi}3-\sin\left(\dfrac{5\pi}3+\dfrac\pi3\right) - \sin0+\sin\dfrac\pi3=0$

So the average value of $f$ over this region is 0.

b. The area of $D$ is $\dfrac{2\pi}3\cdot\dfrac{7\pi}6=\dfrac{7\pi^2}9$ . The integral of $f$ is

$\displaystyle\int_0^{7\pi/6}\int_0^{2\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^{7\pi/6}\left(\cos y-\cos\left(y+\frac{2\pi}3\right)\right)\,\mathrm dy$

$=\sin\dfrac{7\pi}6-\sin\left(\dfrac{7\pi}6+\dfrac{2\pi}3\right) - \sin0+\sin\dfrac{2\pi}3=\dfrac{\sqrt3}2$

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4 0

2 years ago