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2 years ago

22 51/100 as a decimal

Mathematics

1 answer:

Brrunno [24]2 years ago

8 0

22 51/100 as a decimal = 22.51
hope it helps

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Can you find 2/3 of 120

polet [3.4K]

The only thing you should do is to multiply 2/3 by 120 so you will have 2*120/3 = 80 :)))
i hope this is helpful
have a nice day

6 0

2 years ago

Which ordered pair is the solution to the system of equations?

SOVA2 [1]

To find the solution, we use the substitution method.

x+y=-1
x-3y=11

x+y = -1-y
-y

x = -1-y

Now apply the value of x into the other equation.

x-3y=11
-1-y-3y = 11

Combine like terms

-4y -1 = 11
+1 +1
-4y = 12

-4y = 12
-4y/-4 = 12/-4
y = -3

Now, apply the value of Y to one equation to find x.

y = -3

x -3 = -1
+3 +3

x= 2

Now we have the value for both, x and y.
x = 2
y =-3

Final answer: A. (2, −3)

3 0

2 years ago

Factorise p^2 + 30p

Amanda [17]

Answer:

Step-by-step explanation:

hello :

p²+30p = p(p+30)

7 0

3 years ago

Find a · b. |a| = 80, |b| = 50, the angle between a and b is 3π/4.

mart [117]

Answer:

$a\cdot b=-2000\sqrt{2}$

Step-by-step explanation:

Given information: |a| = 80, |b| = 50, the angle between a and b is 3π/4.

We need to find the dot product a · b.

The formula of dot product is

$a\cdot b=|a||b|\cos \theta$

where, θ is the angle between a and b.

Substitute the given values in the above formula.

$a\cdot b=(80)(50)\cos (\frac{3\pi}{4})$

$a\cdot b=4000\cos (\pi-\frac{\pi}{4})$

$a\cdot b=-4000\cos (\frac{\pi}{4})$ $[\because \cos (\pi-\theta)=-\cos \theta]$

$a\cdot b=-4000\frac{1}{\sqrt{2}}$

$a\cdot b=-\frac{4000}{\sqrt{2}}$

Rationalize the above equation.

$a\cdot b=-\frac{4000}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$a\cdot b=-\frac{4000\sqrt{2}}{2}$

$a\cdot b=-2000\sqrt{2}$

Therefore, the value of a · b is $a\cdot b=-2000\sqrt{2}$ .

7 0

2 years ago

Read 2 more answers

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

4 0

2 years ago