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NARA [144]
3 years ago
11

the product of -5 and a number is at least -40 What algebraic inequality represents the phrase and what are the solutions?

Mathematics
1 answer:
nikitadnepr [17]3 years ago
5 0

Answer:

-5x ≥ -40  ; x ≤ 8

Step-by-step explanation:

-5x ≥ -40

x ≤ 8

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Factor the polynomial by finding the greatest common factor 25m^7n^3+10m^4n^3-30m^3n^3
ioda

Answer:

5 m ^3 n^3 ( 125 m^ 4 + 2 m − 6 )

Step-by-step explanation:

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3 years ago
Gustavo ha forrado una pared de 12,5 m cuadrados con 50 paneles cuadrados iguales de madera ¿ Cuantos decímetros cuadrados mide
Stella [2.4K]

Answer:

25 dm²

Step-by-step explanation:

La superficie de nuestra pared tiene 12.5 m²

Si dividimos la superficie por la cantidad de paneles cuadrados, obtenemos la superficie de cada panel

12.5 m² /50 = 0.25 m²

Sabemos que la superficie de un cuadrado se calcula como l²

Convertimos los 0.25 m² a dm² → 0.25 m² . 100 dm² / 1m² = 25 dm²

Esa es la medida de cada papel.

Si calculamos la medida de cada lado será:

√ (25 dm²) = 5 dm

En conclusión un cuadrado con 5 dm de lado tiene 25dm² de superficie

Si convertimos a m² = 0.25 m²

0.25 m² es la medida que cubre 1 panel de papel

para cubrir 12.5 m² necesitariamos (12.5  . 1) /0.25 = 50 paneles.

3 0
3 years ago
How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title
kotegsom [21]
Let t=\sqrt y, so that t^2=y, t^4=y^2, and \mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}. Then

\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}

Now let y=\sqrt2\tan z, so that \mathrm dy=\sqrt2\sec^2z\,\mathrm dz. Then

\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C

Transform back to y to get

\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C

and again to get back a result in terms of t.

\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C
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