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3 years ago

Which equation has the same solution as x2 + 4x – 13 = 5?

Mathematics

2 answers:

max2010maxim [7]3 years ago

6 0

Answer:

x=4

Step-by-step explanation:

timama [110]3 years ago

4 0

X = 4 hope this helps

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Which of the lines graphed has a slope of -2 and a y intercept of 3?

Flura [38]

It's out of D or A I can't really see

3 0

2 years ago

Need alot of help please!!!

kirza4 [7]

A=x
B= 2x
C=3x
If you add them up =
6x =180
÷6 both sides
x=30
A=30
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8 0

3 years ago

Read 2 more answers

Find y if 1030404(3y+36) = 1030403(3y+36).

Blababa [14]

Answer:

y= - 12

Step-by-step explanation:

7 0

3 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

3 years ago

Find the truth set of each predicate.

Sedbober [7]

The truth set for each predicate are:

a) d ∈ { -6, -3, -2, -1, 1, 2, 3, 6}

b) d ∈ { 1, 2, 3, 6}

c) x ∈ { [-2, -1] U [1, 2]}

d) x ∈ {-2, -1, 1, 2}

<h3>How to find the truth set of each predicate?</h3>

We only need to find the sets of values such that the statements are true.

We want to find vales of d, such that d is an integer, and 6/d is an integer.

Here the possible values of d will be:

d ∈ { -6, -3, -2, -1, 1, 2, 3, 6}

Which are all the factors of 6, so all these integers divide 6.

b) Same as before, but this time the domain is a positive integer, so now the truth set will be:

d ∈ { 1, 2, 3, 6}

c) We want to find real values of x such that:

1 ≤ x² ≤ 4

If we apply the square root to the 3 sides, we get two inequalities:

-√1 ≥ x ≥ -√4

√1 ≤ x ≤ √4

Simplifying:

-1 ≥ x ≥ -2

1 ≤ x ≤ 2

So the truth set is:

x ∈ { [-2, -1] U [1, 2]}

c) Same as before, but now we only have integer solutions, so the truth set is:

x ∈ {-2, -1, 1, 2}

If you want to learn more about predicates:

brainly.com/question/18152046

#SPJ1

5 0

2 years ago