What is the slope of a line with points (3, 12)(7, 16)?

If the number 38,469 is rounded to the nearest thousand, how many zeros does the rounded number have?

sladkih [1.3K]

3 because u keep the number Wich is 8 the same but change the three to four

8 0

2 years ago

The question is about trigonometry solving two sides it is a right triangle hypotenuse is X left side is 20 bottom side is 21 i

elena55 [62]

Answer:

x = 29

Angle A = 46.4°

Angle B = 43.6°

Step-by-step explanation:

Since it is a right angle triangle, we can find x from pythagoras theorem;

x = √(20² + 21²)

x = √(400 + 441)

x = √841

x = 29

Using trigonometric ratios we can find angles A and B.

Opposite is 20

Adjacent is 21

Thus; opp/adj = tan B

20/21 = tan B

B = tan^(-1) 0.9524

B = 43.6°

Sum of angles in a triangle is 180°. Thus;

A = 180 - (90 + 43.6)

A = 46.4°

4 0

1 year ago

Draw the image of \triangle ABC△ABCtriangle, A, B, C under a dilation whose center is AAA and scale factor is \dfrac{1}{4} 4 1

Phantasy [73]

Answer:

THE FIRST OPTION

Step-by-step explanation:

GIVE ME BRAINLIEST

3 0

2 years ago

Square had a radius of 5 square root 2 find the apothem

Karo-lina-s [1.5K]

Answer:

Apothem = 5 inches

(circum)radius = 5*sqrt(2) inches

Step-by-step explanation:

6 0

2 years ago

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

$\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}$

Now, we need to express the quadratic polynomials using their roots, as follows:

$ay^2+by+c=a(y-y_1)(y-y_2)$

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}$

Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

Substituting into the rational expression and simplifying:

$\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}$

8 0

4 months ago