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2 years ago

Smaller angle please

Mathematics

2 answers:

TEA [102]2 years ago

6 0

Your answer is 50° maybe

kicyunya [14]2 years ago

4 0

It Will be 50 that the right answer

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Which number are 3 units from -6

Pachacha [2.7K]

There are 2 numbers that are 3 units from -6, it can be either from the left, in that case it would be -9, or to the right would be -3.

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2 years ago

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Irina18 [472]

Answer: Say what is this

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2 years ago

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Which set of side lengths create a right triangle ?

I am Lyosha [343]

Answer:

(8, 15 and 17)

Step-by-step explanation:

$\sqrt{10^{2} +28^{2} } \neq 29$

$\sqrt{6^{2} +8^{2} } \neq 13$

$\sqrt{7^{2} +12^{2} } \neq 14$

$\sqrt{8^{2} +15^{2} } =17$

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2 years ago

Two basketballs are thrown along different paths. Determine if the basketballs’ paths are parallel to each

Paul [167]

Answer:

Since the slopes of the two equations are equivalent, the basketballs' paths are parallel.

Step-by-step explanation:

Remember that:

Two lines are parallel if their slopes are equivalent.
Two lines are perpendicular if their slopes are negative reciprocals of each other.
And two lines are neither if neither of the two cases above apply.

So, let's find the slope of each equation.

The first basketball is modeled by:

$\displaystyle 3x+4y=12$

We can convert this into slope-intercept form. Subtract 3<em>x</em> from both sides:

$4y=-3x+12$

And divide both sides by four:

$\displaystyle y=-\frac{3}{4}x+3$

So, the slope of the first basketball is -3/4.

The second basketball is modeled by:

$-6x-8y=24$

Again, let's convert this into slope-intercept form. Add 6<em>x</em> to both sides:

$-8y=6x+24$

And divide both sides by negative eight:

$\displaystyle y=-\frac{3}{4}x-3$

So, the slope of the second basketball is also -3/4.

Since the slopes of the two equations are equivalent, the basketballs' paths are parallel.

3 0

2 years ago

Simplify completely: the quantity 10 times x to the 6th power times y to the third power plus 20 times x to the third power time

eimsori [14]

Correct answer is D.

$\frac{10x^6y^3 + 20x^3y^2}{5x^3y}
\\
\\ \text{common in the numerator is }
\\
\\ 10x^3y^2
\\
\\ \text{so}
\\
\\ \frac{10x^3y^2(x^3y + 2)}{5x^3y}
\\
\\ \text{cancel common factors... }
\\
\\ 2y(x^3y + 2) = 2x^3y^2 + 4y$

6 0

2 years ago