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valentina_108 [34]
3 years ago
13

Dakota mixes two cereals for breakfast. One gram of cereal A has 6 milligrams of sodium and 7 calories. One gram of cereal B has

8 milligrams of sodium and 5 calories. She wants the bowl of cereal to contain no more than 200 milligrams of sodium and 150 calories. Let "a" equal the number of grams of cereal A and "b" equal the number of grams in cereal B, where are "a" > 0 and "b" > 0. Which system of inequalities represents the possible ways Dakota can mix the cereals.
Mathematics
1 answer:
likoan [24]3 years ago
5 0

Number of grams of cereal A = a

Number of grams of cereal B = b

One gram of Cereal A has: 6 mg of sodium and 7 calories

One gram of Cereal A has: 8 mg of sodium and 5 calories

We need to make sure that the cereal contains maximum 200 mg sodium

⇒ Number of grams of cereal A × Sodium quantity in a gram of cereal A + Number of grams of cereal B × Sodium quantity in a gram of cereal B <= 200 mg

Substituting the required values

⇒ a×6 + b×8 <=200

Further simplifying the above equation

⇒ a×3 + b×4 <=100

Further, we need to make sure that the cereal contains maximum 150 calories

⇒ Number of grams of cereal A × Calories in a gram of cereal A + Number of grams of cereal B × Calories in a gram of cereal B <= 150

⇒ a×7 + b×5 <=150

Hence, the system of inequalities that represents the possible mix is:

a×3 + b×4 <=100 AND a×7 + b×5 <=150

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A multiplier of 1.641 corresponds to a percentage increase or decrease of what percent?
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Answer:

A multiplier of 1.641 corresponds to a percentage increase or decrease of 64.1%.

Step-by-step explanation:

To determine a percentage increase or decrease of a certain number, a calculation consisting of the multiplication of said number by 1 plus the decimals corresponding to the percentage to be increased must be performed. Thus, for example, if you want to increase a number by 50%, you must multiply that number by 1.50. Therefore, if you multiply a number by 1,641, this implies that you are wanting to increase or decrease the initial number by 64.1%.

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3 years ago
Trouble finding arclength calc 2
kiruha [24]

Answer:

S\approx1.1953

Step-by-step explanation:

So we have the function:

y=3-x^2

And we want to find the arc-length from:

0\leq x\leq \sqrt3/2

By differentiating and substituting into the arc-length formula, we will acquire:

\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+4x^2} \, dx

To evaluate, we can use trigonometric substitution. First, notice that:

\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+(2x)^2} \, dx

Let's let y=2x. So:

y=2x\\dy=2\,dx\\\frac{1}{2}\,dy=dx

We also need to rewrite our bounds. So:

y=2(\sqrt3/2)=\sqrt3\\y=2(0)=0

So, substitute. Our integral is now:

\displaystyle S=\frac{1}{2}\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy

Let's multiply both sides by 2. So, our length S is:

\displaystyle 2S=\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy

Now, we can use trigonometric substitution.

Note that this is in the form a²+x². So, we will let:

y=a\tan(\theta)

Substitute 1 for a. So:

y=\tan(\theta)

Differentiate:

y=\sec^2(\theta)\, d\theta

Of course, we also need to change our bounds. So:

\sqrt3=\tan(\theta), \theta=\pi/3\\0=\tan(\theta), \theta=0

Substitute:

\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{1+\tan^2(\theta)}\sec^2(\theta) \, d\theta

The expression within the square root is equivalent to (Pythagorean Identity):

\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{\sec^2(\theta)}\sec^2(\theta) \, d\theta

Simplify:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta

Now, we have to evaluate this integral. To do this, we can use integration by parts. So, let's let u=sec(θ) and dv=sec²(θ). Therefore:

u=\sec(\theta)\\du=\sec(\theta)\tan(\theta)\, d\theta

And:

dv=\sec^2(\theta)\, d\theta\\v=\tan(\theta)

Integration by parts:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\tan^2(\theta)\sec(\theta)} \, d\theta)

Again, let's using the Pythagorean Identity, we can rewrite tan²(θ) as:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^2(\theta)-1)\sec(\theta)} \, d\theta)

Distribute:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^3(\theta)-\sec(\theta)} \, d\theta)

Now, let's make the single integral into two integrals. So:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta-\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)

Distribute the negative:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta

Notice that the integral in the first equation and the second integral in the second equation is the same. In other words, we can add the second integral in the second equation to the integral in the first equation. So:

\displaystyle 2S= 2\int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta

Divide the second and third equation by 2. So: \displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)

Now, evaluate the integral in the second equation. This is a common integral, so I won't integrate it here. Namely, it is:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta))

Therefore, our arc length will be equivalent to:

\displaystyle 2S=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}

Divide both sides by 2:

\displaystyle S=\frac{1}{4}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}

Evaluate:

S=\frac{1}{4}((\sec(\pi/3)\tan(\pi/3)+\ln(\tan(\pi/3)+\sec(\pi/3))-(\sec(0)\tan(0)+\ln(\tan(0)+\sec(0))

Evaluate:

S=\frac{1}{4}((2\sqrt3+\ln(\sqrt3+2))-((1)(0)+\ln(0+1))

Simplify:

S=\frac{1}{4}(2\sqrt 3+\ln(\sqrt3+2)}

Use a calculator:

S\approx1.1953

And we're done!

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Sara made an error in solve the one-step equation below.5x + 9 = 29-9 -9- 4x = 29/4. /4x = -7.25What is the error that Sara made
MAXImum [283]
5x+9=29

To solve this one-step equation you:

1. Substract 9 in both sides of the equation:

In this step is the mistake as Sara gets as result -4x=29 and the corret result of substract 9 in both sides of the equation is 5x=20.

2. Divide both sides of the equation into 5:

Then, the answer that Sara should found for the equation is x=4

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