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3 years ago

11

without carrying a tank bottoms has the size of 3 to 6 m. Find out how many cans of paint of 2 kg are required to paint the insi

de and outside of the tank knowing that 6 square 1 kg

2 answers:

RUDIKE [14]3 years ago

7 0

I think its 3.33 not for sure

Andrews [41]3 years ago

5 0

It is 33.33 hope it helps

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F(x) = 2x-(x+6) for f(-3)

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Plug in -3 for x
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f(-3) = -9
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In how many distinct ways can the letters of the word SEEDS be arranged?

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A box is x to inches high,(2x+3)inches wide, and (2x+5) of inches long. In terms of x, what is the volume (v) of the box?

Advocard [28]

Answer:

The volume of the box is $V=(4x^3+16x^2+15x)\ in^3$

Step-by-step explanation:

we know that

The volume of the box is equal to

$V=LWH$

where

L is the length of the box in inches

W is the width of the box in inches

H is the height of the box in inches

we have

$L=(2x+5)\ in\\W=(2x+3)\ in\\H=x\ in$

substitute in the formula

$V=(2x+5)(2x+3)x$

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3 years ago

find the value of the trigonometric function sin (t) if sec t = -4/3 and the terminal side of angle t lies in quadrant II

kodGreya [7K]

Answer:

$sin(t) =\frac{\sqrt{7}}{4}$

Step-by-step explanation:

By definition we know that

$sec(t) = \frac{1}{cos(t)}$

and

$cos ^ 2(t) = 1-sin ^ 2(t)$

As $sec(t) = -\frac{4}{3}$

Then

$sec(t) = -\frac{4}{3}\\\\\frac{1}{cos(t)} =-\frac{4}{3}\\\\cos(t) = -\frac{3}{4}$

Now square both sides of the equation:

$cos^2(t) = (-\frac{3}{4})^2$

$cos^2(t) = \frac{9}{16}\\\\$

$1-sin^2(t) =\frac{9}{16}\\\\sin^2(t) =1-\frac{9}{16}\\\\sin^2(t) =\frac{7}{16}\\\\sin(t) =\±\sqrt{\frac{7}{16}}$

In the second quadrant sin (t) is positive. Then we take the positive root

$sin(t) =\sqrt{\frac{7}{16}}$

$sin(t) =\frac{\sqrt{7}}{4}$

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3 years ago

What is the ordered pair for 16x – 2y = 74 and 2x – 2y = 4?

34kurt

The answer is ....................
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