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FromTheMoon [43]
3 years ago
13

HELPP QUICKK PLEASSEEE​

Mathematics
1 answer:
aleksklad [387]3 years ago
3 0

Answer:

Questions are blur can you post a much clearer one?

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Will give brainliest!!
swat32

Answer:

its 5 hours i think

Step-by-step explanation:

3 0
3 years ago
You are going to mix a gallon bucket of window cleaner. The instructions direct you to mix 1 part cleaner to 3 parts water. How
Pie
One gallon equals to 4 quarts.

The instructions tell you to add 1 part of cleaner and 3 parts of water, in total 4 parts. It makes it easy as you can take that one part = one quart.

You will use 3 quarts of water and 1 quart of cleaner. The answer is D.


4 0
3 years ago
Read 2 more answers
Plz help with questions 12 box and 13-14 Due tomorrow !!!
ivann1987 [24]
12. 6$
13. 60 birds
14. 10 inch
8 0
3 years ago
N.
noname [10]

Answer:

287.1 inches of the canvas.

Step-by-step explanation:

To solve this, we need to first figure out the total area of the canvas. To do that, multiply width by height.

29*33=957

Now set up your equation for solving for the area of the canvas that the rose covers.

x/957=30/100

We did it where: x is the area of the rose covers, 957 is the amount of inches that the canvas takes up, and the right side of the equation is the percent.

Now cross multiply.

100x=28,710

Now divide both sides by 100.

x=287.1

The red rose covers 287.1 inches of the canvas.

6 0
3 years ago
Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering
Svetllana [295]

Answer:

L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x

Step-by-step explanation:

We are asked to find the tangent line approximation for f(x)=\sqrt{10+x} near x=0.

We will use linear approximation formula for a tangent line L(x) of a function f(x) at x=a to solve our given problem.

L(x)=f(a)+f'(a)(x-a)

Let us find value of function at x=0 as:

f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}

Now, we will find derivative of given function as:

f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}

f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)

f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1

f'(x)=\frac{1}{2\sqrt{10+x}}

Let us find derivative at x=0

f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}

Upon substituting our given values in linear approximation formula, we will get:

L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)  

L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0

L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x

Therefore, our required tangent line for approximation would be L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x.

8 0
3 years ago
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