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3 years ago

Mn+2n+n^2+2m=(n+2)() = ?

Mathematics

1 answer:

sattari [20]3 years ago

3 0

$mn + 2n + {n}^{2} + 2m \\ = (n + 2)(m + n)$

Hope this helps. - M

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PY=2, HP=3, find HY<br><br> Geometry

klemol [59]

Answer:

$HY=\sqrt{5}\ units$

Step-by-step explanation:

we know that

The triangle HPY is a right triangle (because the angle ∠HYP is a right angle)

Applying the Pythagorean Theorem

$HP^2=HY^2+PY^2$

substitute the given values

$3^2=HY^2+2^2$

Solve for HY

$9=HY^2+4$

$HY^2=9-4$

$HY^2=5$

$HY=\sqrt{5}\ units$

5 0

3 years ago

8. Mark chose a number between 0.437 and 0.436 and multiplied it by 4. Then, he

Dvinal [7]

Answer:

y=¾(4x-20)

y=3x-15

Final answer= y - 3(x+12)

=y - 3x-36

=3x-15-3x-36

= -51

5 0

3 years ago

Read 2 more answers

-/2 points

Rom4ik [11]

Answer:

Part 1) The domain of the quadratic function is the interval (-∞,∞)

Part 2) The range is the interval (-∞,1]

Step-by-step explanation:

we have

$f(x)=-x^2+14x-48$

This is a quadratic equation (vertical parabola) open downward (the leading coefficient is negative)

step 1

Find the domain

The domain of a function is the set of all possible values of x

The domain of the quadratic function is the interval

(-∞,∞)

All real numbers

step 2

Find the range

The range of a function is the complete set of all possible resulting values of y, after we have substituted the domain.

we have a vertical parabola open downward

The vertex is a maximum

Let

(h,k) the vertex of the parabola

The range is the interval

(-∞,k]

Find the vertex

$f(x)=-x^2+14x-48$

Factor -1 the leading coefficient

$f(x)=-(x^2-14x)-48$

Complete the square

$f(x)=-(x^2-14x+49)-48+49$

$f(x)=-(x^2-14x+49)+1$

Rewrite as perfect squares

$f(x)=-(x-7)^2+1$

The vertex is the point (7,1)

therefore

The range is the interval

(-∞,1]

6 0

3 years ago

Need some help with this one please

Airida [17]

Answer: Choice A

y = (-3/4)(x + 4) + 6

=====================================================

Let's go through the answer choices

Choice A is something we'll come back to
Choice B is false because the line does not go uphill as we move from left to right. The graphed line has a negative slope, which contradicts what choice B is saying.
Choice C is false for similar reasons as choice B. The slope should be negative.
Choice D has a negative slope, but the y intercept is wrong. The y intercept should be 3. So choice D is false as well.

We've eliminated choices B through D.

Choice A must be the answer through process of elimination.

------------

Here's an alternative method:

If we started at a point like (0,3) and move to (4,0), note how the slope is -3/4

This is because we've moved down 3 units and to the right 4 units.

m = slope = rise/run = -3/4

We can also use the slope formula m = (y2-y1)/(x2-x1) to see this.

Then we pick on a point that is on the diagonal line. It could be any point really, but the point your teacher used for choice A is (x1,y1) = (-4,6)

So,

y - y1 = m(x - x1)

y - 6 = (-3/4)(x - (-4))

y - 6 = (-3/4)(x + 4)

y = (-3/4)(x + 4) + 6

7 0

4 years ago

Solve for A ? <br><br> Anyone willing to help me :)

GrogVix [38]

Answer:

Step-by-step explanation:

Use the cosine law:

$BC^2=AB^2+AC^2-2(AB)(AC)\cos(\angle A)$

We have:

$BC=a\\\\AB=4\\\\AC=3\\\\m\angle A=32^o\to\cos32^o\approx0.848$

Substitute:

$a^2=4^2+3^2-2(4)(3)(0.848)\\\\a^2=16+9-20.352\\\\a^2=4.648\to a=\sqrt{4.648}\\\\a\approx2.2$

5 0

3 years ago