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Simplify the product using the distributive property. (2h-3)(3h+4)

myrzilka [38]

6h^2-h-12 apply the foil method : (a+b)(c+d)=ac+ad+bc+bd

4 0

3 years ago

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HELP ME PLEASE <br> solve the right triangle

Amiraneli [1.4K]

ST = 7.2

RT = 4.04

m∠S = 34°

Solution:

Given ΔRST is a right triangle.

RS = 6, ∠T = 56°

$$\csc A=\frac{\text {Hypotenuse} }{\text {Opposite side}}$

$$\csc A=\frac{ST}{RS}$

$$\csc 56^\circ=\frac{ST}{6}$

$$\csc 56^\circ \times 6=ST$

1.206 × 6 = ST

7.236 = ST

ST = 7.2

$$\cot A=\frac{\text {Adjacent side} }{\text {Opposite side}}$

$$\cot A=\frac{RT}{RS}$

$$\cot 56^\circ=\frac{RT}{6}$

$$\cot 56^\circ \times 6=RT$

0.674 × 6 = RT

4.044 = RT

RT = 4.04

To find the measure of angle S.

Sum of the interior angles of the triangle = 180°

m∠R + m∠S + m∠T = 180°

90° + m∠S + 56° = 180°

m∠S + 146° = 180°

m∠S = 180° – 146°

m∠S = 34°

8 0

3 years ago

Let us have four distinct collinear points $a,$ $b,$ $c,$ and $d$ on the cartesian plane. the point $c$ is such that $\dfrac{ab}

Deffense [45]

Start with a line segment connecting two points, A and B. $\dfrac{DA}{BA}=3$ means DA is 3 times longer than BA. Clearly, D cannot fall between A and B because that would mean DA is shorter than BA. So there are two possible locations where D can be placed on the line relative to A and B.

But with $\dfrac{DB}{BA}=2$ , or the fact that DB is 2 times longer than BA, we can rule out one of these positions; referring to the attachment, if we place D to the left of A, then DB would be 4 times longer than BA.

Finally, $\dfrac{AB}{CB}=\dfrac12$ , so that CB is 2 times longer than AB. Again we have two possible locations for point C (it cannot fall between A and B), but one of them forces C to occupy the same point as D. However, A, B, C, D are distinct, so C must fall to the left of A.

Now let $d$ be the length of AB. Then the length of CD in terms of $d$ is $4d$ . We have the coordinates of C and D, and the distance between them is $\sqrt{(4-0)^2+(0-4)^2}=4\sqrt2$ . So

$4d=4\sqrt2\implies d=\sqrt2$

The slope of the line through C and D is

$\dfrac{0-4}{4-0}=-1$

and so the equation of the line through these points is

$y-4=-(x-0)\implies x+y=4$

So the coordinates of A are $(x,y)=(x,4-x)$ . The distance between C and A is $d=\sqrt2$ , so we have

$\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1$

Since A falls to the right of C (in the $x,y$ plane, not just in the sketch), we know to take the positive value $x=1$ . Then the $y$ coordinate is $y=4-1=3$ .