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3 years ago

A number increased by 3 is 19

Mathematics

2 answers:

nydimaria [60]3 years ago

4 0

Answer: 16

steps: 19-3=16

Hope this helps u :)

navik [9.2K]3 years ago

3 0

Answer:

The number would be 16.

Step-by-step explanation:

If you have a specific number of apples, and then pluck 3 more, you have 19. So this can be modeled as 19-3, which gives you the original

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Eighty-eight less than m is smaller than -176.<br> Answer in interval notation

IrinaK [193]

Answer:

(- ∞ , 88)

Step-by-step explanation:

1) m - 88 < - 176

add 88 to both sides

2) m < -88

3) interval notation: ( - ∞, 88)

8 0

3 years ago

What’s the answer

Dmitriy789 [7]

The y intercept is 3

4 0

3 years ago

Solve for c.<br> a(c+b) = d.

kipiarov [429]

Answer:

Step-by-step explanation:

Hello,

we assume that a is different from 0 otherwise there is not much we can say on c and then we can write

a(c+b)=d

<=> ac+ab=d

<=> ac = d - ab

<=> $c = \dfrac{d-ab}{a} =\dfrac{d}{a}-b$

Hope this helps

3 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

HELP ME WITH THIS PLEASE

olasank [31]

Answer:

use photomath

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers