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Olegator [25]
3 years ago
13

How many times can 5 go into 7

Mathematics
2 answers:
VikaD [51]3 years ago
4 0
Once because 5 times 1 is 5 and that's closest it will go into Hope that helped Have a good day! :-)
Basile [38]3 years ago
3 0
Once because 7 divided by 5 is equal to 1.2 so there is a remainder
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2 years ago
Write the ratios for sin X and cos X. Given that sin is opposite over hypotenuse and
Romashka-Z-Leto [24]
The trigonometric ratios are used for right angle triangle.The basic rule to followed is SOH CAH TOA

SOH means Sinx=opposite/Hyp.
CAH means Cosx = adj./Hyp.
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3 years ago
Matt is playing a game. He gains 7 points, loses 10 points, and then loses 8 points. What is his final score? *​
Juli2301 [7.4K]

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Step-by-step explanation:

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8 0
2 years ago
Read 2 more answers
Linn is making cookies. Each batch of cookies
Pachacha [2.7K]

Answer:

4 cups of sugar require 14 cups of sugar

Step-by-step explanation:

Given

\frac{7}{8}\ cup\ flour = \frac{1}{4}\ cup\ sugar

Required

Determine the cups of flour for 4 cups of sugar

First, we need to determine the unit rate for 1 cup of sugar.

This is done by multiplying both sides by 4

4 * \frac{7}{8}\ cup\ flour = \frac{1}{4}\ cup\ sugar * 4

\frac{7}{2}\ cup\ flour = 1\ cup\ sugar

Next, we determine cup of flour for 4 cups of sugar.

This is done by multiplying both sides by 4

4 * \frac{7}{2}\ cup\ flour = 1\ cup\ sugar * 4

2 * 7\ cup\ flour = 4\ cup\ sugar

14\ cup\ flour = 4\ cup\ sugar

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8 0
3 years ago
Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?
sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have V=IR^{2} and let H be the subset of V of all points on the line

4x+3y=12

We need to find if H is a subspace of the vector space V.

In IR^{2} all the possibilities for own subspace of the vector space IR^{2} are :

  • IR^{2} itself.
  • The vector 0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]
  • All lines in IR^{2} that passes through the origin  (  0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]  )

We know that H is the subset of IR^{2} of all points on the line 4x+3y=12

If we look at the equation, the point \left[\begin{array}{c}0&0\end{array}\right] doesn't verify it because :

4x+3y=12\\4(0)+3(0)=12\\0=12

Which is an absurd. Therefore, H doesn't contain the origin (and H is a line in IR^{2}). Finally, it can't be a vector space of V=IR^{2}

8 0
3 years ago
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