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zaharov [31]
3 years ago
10

Name a ray .please help me

Mathematics
2 answers:
NNADVOKAT [17]3 years ago
6 0
A ray is part of a line with one endpoint and goes on in one direction.
Dafna1 [17]3 years ago
4 0
A ray is a line with an end hope this helps

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nirvana33 [79]
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juan needs to measure 6 cups of flour for a recipe.he only has a 1/4 cup measuring cup.how many times must he fill the measuring
Luba_88 [7]

Answer:

Juan would need to fill up the 1/4 measuring cup 24 times.

Step-by-step explanation:

So to measure one cup of flour with a 1/4 measuring cup, you would need to fill up the 1/4 measuring cup 4 times. Since we want 6 cups, you would need to fill up the 1/4 measuring cup 24 times. 6 times 4 = 24. Hope this helps!

7 0
3 years ago
If I have 24 jolly ranchers and 64 starbursts what is the largest amount of bags I can fill? How many starbursts will be I’m eac
Rus_ich [418]

Answer:

Largest amount of bags that can be filled: 11

Starbursts in each bag: 3

Step-by-step explanation:

Find the greastest common factor between these two (which is 8) and divide the sum of the jolly ranchers and starbursts-which will be 88-by the greatest common factor. This will leave you with 11 bags that can be filled with 8 candies. To find the starbursts, divide 24 by 8 (3.)

5 0
3 years ago
Two-stroke fuel for a lawn mower requires an oil: petrol ratio of 1:40. How much petrol is added to 50ml of oil
bearhunter [10]
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8 0
3 years ago
Solve the following equation. X cubed minus 6X squared plus 6X equals zero
anyanavicka [17]

We have to solve this equation:

x^3-6x^2+6x=0

Third degree polynomials like this one are not easily solved, but this one has a root at x = 0. The let us factorize this polynomial as x times a second degree polynomial:

\begin{gathered} x^3-6x^2+6x=0 \\ x(x^2-6x+6)=0 \end{gathered}

Now we can find the roots of the quadratic polynomial as:

\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{6\pm\sqrt[]{36-24}}{2} \\ x=\frac{6\pm\sqrt[]{12}}{2} \\ x=\frac{6\pm\sqrt[]{4\cdot3}}{2} \\ x=\frac{6\pm2\sqrt[]{3}}{2} \\ x=3\pm\sqrt[]{3} \\ x_1=3-\sqrt[]{3} \\ x_2=3+\sqrt[]{3} \end{gathered}

Then, the solutions to the equation are:

x = 0

x = 3 - √3

x = 3 + √3

4 0
1 year ago
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