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skad [1K]
3 years ago
15

Problem PageQuestion Michael drives a truck for a soft drink company. His truck is filled with 15-ounce cans and 70-ounce bottle

s. Let c be the number of 15-ounce cans the truck is carrying, and let b be the number of 70-ounce bottles. The truck must be carrying less than 4000 pounds (64,000 ounces). Using the values and variables given, write an inequality describing this.
Mathematics
1 answer:
Colt1911 [192]3 years ago
6 0

Answer:345

Step-by-step explanation:

bvteswasdftgyhujiuytrewsdrfgthyhtrdwdefrtyujtrewqsdefrtgyhjuijytrewsqwdefrgtyuijrvdswdefrgth

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5 to the power of what equals 60
ycow [4]
<h2>8.6736174e+41 is going to be your answer hope it not to late for your question</h2>
4 0
3 years ago
Can someone help me with this question please!!!!!
faltersainse [42]
From the equation, it looks like we can set up a triangle like this.  One side is x and the other is x + 20 because the northbound car traveled 20 miles farther.  The hypotenuse is 100 because they were 100 miles apart.

Next, we can use the Pythagorean Theorem to solve for x:

a² + b² = c²
x² + (x + 20)² = 100²

This simplifies to:

x² + x² + 40x + 400 = 10000
2x² + 40x = 9600
2x² + 40x - 9600 = 0

After that, we can divided each side by 2:

x² + 20x - 4800 = 0

Next, we have to factor:

(x - 60)(x + 80) = 0
x = 60, -80

Since we know that distance can't be negative, 60 is the valid answer in this case.

x = 60
x + 20 = 80

Using this information, we know that the westbound car traveled 60 miles and the northbound car traveled 80 miles.

6 0
2 years ago
Question 5
sergey [27]

Explanation : If 66 is a factor of this unknown value, then 22 must be as well considering that 22 is a factor off 66. Let's say that this large value is 330. It is a multiple of 66, as 66 * 5 = 330. At the same time 22 * 15 = 330, so 330 is a multiple of 22 as well - or vice versa, 12 is a factor of 330.

We can also tell that 15, 22 fit into 330 through another approach. 22 * 3 = 66, and 66 * 5 = 330, so 5 * 3 = 15 - the same value. This proves that 22 will always be a factor of a value that is the factor of 66.

3 0
3 years ago
Read 2 more answers
What inequalities have the solution set graphed in the number line? Select two options.
Akimi4 [234]
The answer should be 1 and three
7 0
3 years ago
If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
Alexxx [7]
\bf f(x)=y=2x+sin(x)&#10;\\\\\\&#10;inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)&#10;\\\\\\&#10;\textit{now, the "y" in the inverse, is really just g(x)}&#10;\\\\\\&#10;\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\&#10;-----------------------------\\\\

\bf \textit{let's use implicit differentiation}\\\\&#10;1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}&#10;\\\\\\&#10;1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\&#10;-----------------------------\\\\&#10;g'(2)=\cfrac{1}{2+cos[g(2)]}

now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2&#10;\\\\\\&#10;-----------------------------\\\\&#10;g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
5 0
2 years ago
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