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3 years ago

50.) Find the value of b so that 330_b= 226_b+5

Mathematics

1 answer:

lyudmila [28]3 years ago

7 0

First u do 330b - 226b then divide 5 by that. b=0.048

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A point on the terminal side of an acute angle in standard position is (1, 2 √2). Find the cosine value of this angle.

kozerog [31]

Answer:

$cos\theta=1/3$

Step-by-step explanation:

The point at the terminal side of an acute angle is given by $(1, 2\sqrt{2} )$ .

That is,

$x = 1$ and $y= 2\sqrt{2}$ .

Let r be the length of line segment drawn from the origin to the point and is given by the formula:

$r = \sqrt{x^{2} + y^{2}}$

Substituting the values of x and y into r,

$r = \sqrt{1^{2} + (2\sqrt{2} )^{2}}$

$r = \sqrt{1 + 8}$

$r = \sqrt{9}$

Thus, $r=3$

Also, $cos\theta$ is given by:

$cos\theta = x/r\\$

Substituting values of x and r,

$cos\theta = 1/3\\$

4 0

3 years ago

X + 5 > 7<br> it’s inequalities

Hatshy [7]

Answer:

X>2

Step-by-step explanation:

X+5>7

X>2

7 0

3 years ago

Read 2 more answers

A box with a square base and open top must have a volume of 296352 c m 3 . We wish to find the dimensions of the box that minimi

mestny [16]

Answer:

Base Length of 84cm
Height of 42 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 296352 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume, $V=x^2h=296352$

$h=\dfrac{296352}{x^2}$

Surface Area of the box = Base Area + Area of 4 sides

$A(x,h)=x^2+4xh\\$Substitute h=\dfrac{296352}{x^2}\\A(x)=x^2+4x\left(\dfrac{296352}{x^2}\right)\\A(x)=\dfrac{x^3+1185408}{x}$

Step 2: Find the derivative of A(x)

$If\:A(x)=\dfrac{x^3+1185408}{x}\\A'(x)=\dfrac{2x^3-1185408}{x^2}$

Step 3: Set A'(x)=0 and solve for x

$A'(x)=\dfrac{2x^3-1185408}{x^2}=0\\2x^3-1185408=0\\2x^3=1185408\\$Divide both sides by 2\\x^3=592704\\$Take the cube root of both sides\\x=\sqrt[3]{592704}\\x=84$

Step 4: Verify that x=84 is a minimum value

We use the second derivative test

$A''(x)=\dfrac{2x^3+2370816}{x^3}\\$When x=84$\\A''(x)=6$

Since the second derivative is positive at x=84, then it is a minimum point.

Recall:

$h=\dfrac{296352}{x^2}=\dfrac{296352}{84^2}=42$

Therefore, the dimensions that minimizes the box surface area are:

Base Length of 84cm
Height of 42 cm.

5 0

3 years ago

Help pleaseeeeeeeeee

storchak [24]

I am pretty sure that the answer is the first one but I’m not 100%

8 0

3 years ago

Read 2 more answers

Best explained and correct answer gets brainliest!

sertanlavr [38]

Same side, interior angles of parallel lines cut by a transversal are supplementary.
The sum of the measures of supplementary angles is 180.

Angles 4 and 6 are same side, interior angles, so their measures add to 180.

m<4 + m<6 = 180

109 + m<6 = 180

m<6 = 180 - 109

m<6 = 71

Answer: m<6 = 71 deg.

8 0

3 years ago