Solve -5.55-8.55c+4.35c for me please

Mathematics

1 answer:

serious [3.7K]3 years ago

7 0

Answer:

− 5.55 − 4.2 c

Step-by-step explanation:

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Expand each expression. Then simplify the expression.<br><br> 7(5 + 4w) + 6(8w - 3)

Ilia_Sergeevich [38]

Answer:

76w + 17

Step-by-step explanation:

7 ( 5 + 4w ) + 6 ( 8w - 3 )

= 7 ( 5 ) + 7 ( 4w ) + 6 ( 8w ) - 6 ( 3 )

= 35 + 28w + 48w - 18

= 28w + 48w + 35 - 18

= 76w + 17

3 0

2 years ago

Oduct of (x2 + 3)(x - 5).

Zanzabum

Answer: 2x^2-7x-15

Step-by-step explanation:

reorder the terms (x*2+3)*(x-5)

multiply the parentheses

(2x+3)*(x-5)

collect like terms (2x^2-10x+3x-15)

product= 2x^2-7x-15

8 0

2 years ago

A projectile is launched from the ground at an angle of theta above the horizontal with an initial speed of v in ft/s. The rang

VARVARA [1.3K]

Answer:

$\theta=24.4^\circ$

Step-by-step explanation:

The formula you wrote has the fraction the other way around. You can find that the range of a projectile is in reality approximated by the equation $x=\frac{v^2}{g} sin(2\theta)=\frac{v^2}{32ft/s^2} sin(2\theta)$ , where we will use ft for distances.

From the given equation we have then $\frac{x32ft/s^2}{v^2} =sin(2\theta)$ , which means $\theta=\frac{Arcsin(\frac{x32ft/s^2}{v^2})}{2}$ since the arcsin is the inverse function of the sin.

Since we have x = 170 ft and v = 85 ft/s, we can substitute these values from the equation written, and we will have $\theta=\frac{Arcsin(\frac{(170ft)(32ft/s^2)}{(85ft/s)^2})}{2}$ , and from now on we have just to use a calculator, obtaining $\theta=\frac{Arcsin(0.75294117647)}{2}=\frac{48.84579804^\circ}{2}=24.4^\circ$