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2 years ago

Compare A and B if 70% of A is equal to $0.7a and 35% of B is equal to $0.35a

Mathematics

2 answers:

MrRa [10]2 years ago

3 0

Answer:

A=B

Step-by-step explanation:

Equation 1: 70% of A = 0.7a

Equation 2: 35% of B = 0.35a

Equation 3 = (Equation 2) • 2

Equation 3: 70% of B = 0.7a

Equation 3 = Equation 1

A = B

pogonyaev2 years ago

3 0

Answer:

A=B

becuase it is and ya

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Can someone help with this

Anit [1.1K]

Answer: Draw a line through the two points (0,2) and (3,3). See the attached image below.

How did I get these two points? The first point comes from the fact that "at the start, the height is 2 inches" meaning that x = 0 hours and y = 2 inches pair up together. This is show as point A. This is the y intercept as this is the location where the blue graph crosses the vertical y axis.

Point B is found by moving 1 unit up from point A and 3 units to the right to land on (3,3). So we have x = 3 hours and y = 3 inches pair up together.

Note: the slope is rise/run = 1/3 helping us determine the unit rate or speed in which the water is rising (ie the water is rising one-third of an inch per hour)

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3 years ago

General Electric Company has an assembly line where light bulbs are produced. The ratio of defective bulbs to good bulbs produce

kati45 [8]

Answer: 200 bulbs will not be defective.

Step-by-step explanation:

The ratio of defective bulbs to good bulbs produced each day is 2 to 10. This ratio can also be expressed as 1 to 5 by reducing to lowest terms.

The total ratio is the sum of the proportions.

Total ratio = 1 + 5 = 6

This means that if n bulbs is produced, the number of defective bulbs would be

1/6 × n

The number of non defective would be

5/6 × n

Since n = 240, then the number of bulbs that will not be defective is

5/6 × 240 = 200 bulbs

4 0

3 years ago

Calculus help with the graph of a integral and finding the following components.

lys-0071 [83]

A)

$\bf \displaystyle F(x)=\int\limits_{4}^{\sqrt{x}}\cfrac{2t-1}{t+2}\cdot dt\qquad x=16\implies \displaystyle F(x)=\int\limits_{4}^{\sqrt{16}}\cfrac{2t-1}{t+2}\cdot dt
\\\\\\
\displaystyle F(x)=\int\limits_{4}^{4}\cfrac{2t-1}{t+2}\cdot dt\implies 0
$

why is 0? well, the bounds are the same.

b)

let's use the second fundamental theorem of calculus, where F'(x) = dF/du * du/dx

$\bf \displaystyle F(x)=\int\limits_{4}^{\sqrt{x}}\cfrac{2t-1}{t+2}\cdot dt\implies F(x)=\int\limits_{4}^{x^{\frac{1}{2}}}\cfrac{2t-1}{t+2}\cdot dt\\\\
-------------------------------\\\\
u=x^{\frac{1}{2}}\implies \cfrac{du}{dx}=\cfrac{1}{2}\cdot x^{-\frac{1}{2}}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}}\\\\
-------------------------------\\\\$

$\bf \displaystyle F(x)=\int\limits_{4}^{u}\cfrac{2t-1}{t+2}\cdot dt\qquad F'(x)=\cfrac{dF}{du}\cdot \cfrac{du}{dx}
\\\\\\
\displaystyle\cfrac{d}{du}\left[ \int\limits_{4}^{u}\cfrac{2t-1}{t+2}\cdot dt \right]\cdot \cfrac{du}{dx}\implies \cfrac{2u-1}{u+2}\cdot \cfrac{1}{2\sqrt{x}}
\\\\\\
\cfrac{2\sqrt{x}}{\sqrt{x}+2}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{2\sqrt{x}-1}{2x+4\sqrt{x}}$

c)

we know x = 16, we also know from section a) that at that point f(x) = y = 0, so the point is at (16, 0), using section b) let's get the slope,

$\bf \left. \cfrac{2\sqrt{x}-1}{2x+4\sqrt{x}} \right|_{x=16}\implies \cfrac{2\sqrt{16}-1}{2(16)+4\sqrt{16}} \implies \cfrac{7}{48}\\\\
-------------------------------\\\\
\begin{cases}
x=16\\
y=0\\
m=\frac{7}{48}
\end{cases}\implies \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-0=\cfrac{7}{48}(x-16)
\\\\\\
y=\cfrac{7}{48}x-\cfrac{7}{3}\implies y=\cfrac{7}{48}x-2\frac{1}{3}$

d)

$\bf 0=\cfrac{2\sqrt{x}-1}{2x+4\sqrt{x}}\implies 0=\cfrac{2\sqrt{x}-1}{(\sqrt{x}+2)(2\sqrt{x})}$

now, we can get critical points from zeroing out the derivative, we also get critical points from zeroing out the denominator, however, the ones from the denominator are points where the function is not differentiable, namely, is not a smooth curve, is a sharp jump, a cusp, or a spike, and therefore those points are usually asymptotic, however, they're valid critical points, let's check both,

$\bf 0=\cfrac{2\sqrt{x}-1}{2x+4\sqrt{x}}\implies 0=\cfrac{2\sqrt{x}-1}{(\sqrt{x}+2)(2\sqrt{x})}\\\\
-------------------------------\\\\
0=2\sqrt{x}-1\implies 1=2\sqrt{x}\implies \cfrac{1}{2}=\sqrt{x}\implies \left(\cfrac{1}{2} \right)^2=x
\\\\\\
\boxed{\cfrac{1}{4}=x}\\\\
-------------------------------\\\\
0=(\sqrt{x}+2)(2\sqrt{x})\implies 0=2\sqrt{x}\implies \boxed{0=x}
\\\\\\
0=\sqrt{x}+2\implies -2=\sqrt{x}\implies (-2)^2=x\implies \boxed{4=x}$

now, doing a first-derivative test on those regions, we get the values as in the picture below.

so, you can see where is increasing and decreasing.

5 0

3 years ago

How could you describe the maximum the LCM between two numbers could be?

Licemer1 [7]

Answer:

the product of the two numbers

Step-by-step explanation:

The LCM of two numbers is their product, divided by their GCF. If their GCF is 1, then the LCM will be their product. That is the most it can be.

We assume you're concerned with whole numbers. For rational numbers, the GCF may be a fraction, so the LCM may be larger than the product of the numbers.

LCM(5, 6) = 5·6/GCF(5, 6) = 5·6/1 = 30

LCM(1/2, 3/4) = (1/2)(3/4)/GCF(1/2, 3/4) = (3/8)/(1/4) = 3/2

3 0

3 years ago

Some help me please and thank you

lisov135 [29]

Answer:

6/15 + 5/15 = 11/15

Step-by-step explanation:

cross multiply the numbers to get a common denominators, then add

5 0

3 years ago