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kirill [66]
3 years ago
6

there are 1,600 kilocalories in the 5cups of dog food that mike gives his adult dog. mike gives his puppy 2cups of the same dog

food. how many kilocalories are there in this 2 cup serving
Mathematics
1 answer:
bija089 [108]3 years ago
6 0
There are 640 kilocalories in the 2 cups of dog food. :)
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kylie opened a bank account. she deposited 76.50 into her account every month for 10 months. she used 30.50 every month to pay f
topjm [15]
The total amount of money Kylie spent to go to talent camp is 115. 

76.5 *10 =765 (Amount of money she deposited in 10 months)
30.5 *10 = 305 (Amount of money she used to pay for singing lessons in 10 months)
765 - 305 = 460 (Amount remaining after singing lessons for 10 months)
460/4 = 115 (Amount she used to go to a talent camp)
5 0
3 years ago
Help please!!!!!! and explanation if possible
loris [4]
The correct answer is 3



the triangle sides are equal. so :
4x - 1 = 2x + 5 \\ 4x - 2x = 1 + 5 \\ 2x = 6 \\ x =  \frac{6}{2}  = 3



good luck
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4 years ago
How long will it take pablo to bike 72 miles if he is biking at a speed of 8 mph
BartSMP [9]

Think as seven (7) hours

3 0
3 years ago
Read 2 more answers
Find the equation of the line tangent to the graph of
garik1379 [7]

Answer:

\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}

Step-by-step explanation:

We want to find the equation of the line tangent to the graph of:

\displaystyle y=\sin^{-1}\big(\frac{x}{5}\big)\text{ at } x=\frac{5}{2}

So, we will find the derivative of our equation first. Applying the chain rule, we acquire that:

\displaystyle y^\prime=\frac{1}{\sqrt{1-(\frac{x}{5})^2}}\cdot\frac{1}{5}

Simplify:

\displaystyle y^\prime=\frac{1}{5\sqrt{1-\frac{x^2}{25}}}

We can factor out the denominator within the square root:

\displaystyle y^\prime =\frac{1}{5\sqrt{\frac{1}{25}\big(25-x^2)}}

Simplify:

\displaystyle y^\prime=\frac{1}{\sqrt{25-x^2}}

So, we can find the slope of the tangent line at <em>x</em> = 5/2. By substitution:

\displaystyle y^\prime=\frac{1}{\sqrt{25-(5/2)^2}}

Evaluate:

\displaystyle y^\prime=\frac{1}{\sqrt{75/4}}=\frac{1}{\frac{5\sqrt{3}}{2}}=\frac{2\sqrt{3}}{15}

We will also need the point at <em>x</em> = 5/2. Using our original equation, we acquire that:

\displaystyle y=\sin^{-1}(\frac{1}{2})=\frac{\pi}{6}

So, a point is (5/2, π/6).

Finally, by using the point-slope form, we can write:

\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}(x-\frac{5}{2})

Distribute:

\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}x+\frac{-\sqrt{3}}{3}

Isolate. Hence, our equation is:

\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}

7 0
3 years ago
How can you solve for the unknown in each equation
creativ13 [48]

Answer:

it depends on the equation on what your solving

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