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2 years ago

Find the product. (x2+6x +9)(x?) - What’s the product

Mathematics

2 answers:

Annette [7]2 years ago

7 0

(x²+6x+9)x=x(x+3)²

.............

lesya692 [45]2 years ago

4 0

Answer:

Step-by-step explanation:

(x² + 6x + 9)(x)= x²*x + 6x*x + 9*x = x³ + 6x² + 9x

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riangle XYZ is translated 4 units up and 3 units left to yield ΔX'Y'Z'. What is the distance between any two corresponding point

tester [92]

Answer:

5 units

Step-by-step explanation:

According to the given statement Δ XYZ is translated 4 units up and 3 units left to yield ΔX'Y'Z' which means that each point in ΔXYZ is moved 4 units up and moved 3 units left.

To find the distance of each corresponding point we will use the Pythagorean theorem which states that the square of the length of the Pythagorean of a right triangle is equal to the sum of the squares of the length of other legs

The square of the required distance = 4^2+3^2 = 16+9 =25

By taking root of 25 we get:

√25 = 5

Thus, we can conclude that the the distance between any two corresponding points on ΔXYZ and ΔX′Y′Z′ is 5 units. ..

7 0

2 years ago

Can anyone explain how to do this???

Alex73 [517]

Answer:

perimeter is length plus length width plus width

area is length times width

Step-by-step explanation:

(x+4)+(2x-23)=perimeter

(x+4)(2x-23)

hope this helps

8 0

2 years ago

Read 2 more answers

The coordinates of polygon ABCD are A(-4,-1), B(-2,3), C(2,2), and D(4,-3). Use these coordinates to complete the sentences belo

oee [108]

Answer:

The perimeter of polygon ABCD, to the nearest thousandth units is

22.227 units

The perimeter of polygon ABCD', to the nearest thousandth, would be 20.980 units

The area of polygon ABCD' is 19.5 units²

Step-by-step explanation:

The coordinates forming the polygon are

A (-4,-1), B(-2,3), C(2,2), and D(4,-3)

The perimeter then is given by the sum of the length of the sides as follows;

Length of line between two X and Y points distance, xi and yj apart is

length XY = $\sqrt{x^2+y^2}$

Therefore, the length between points AB is

length AB = $\sqrt{(-4 - (-2))^2+(-1-3)^2}$

= $\sqrt{(-4 +2)^2+(-4)^2} = \sqrt{-2^2+-4^2} = \sqrt{20}$ = 4.472 units

Similarly, length BC is given by

Length BC = $\sqrt{(-4)^2+1^2} = \sqrt{17}$ = 4.123 units

Length CD = $\sqrt{(-2)^2+5^2} = \sqrt{29}$ = 5.385 units

Length DA = $\sqrt{(8)^2+(-2)^2} = \sqrt{68}$ = 8.246 units

The perimeter is equal to;

length AB + Length BC + Length CD + Length DA

= 4.472 + 4.123 + 5.385 + 8.246 = 22.227 units

If the point D is moved up 2 units and left 1 unit we have

D' = x-1, y+2 where x and y are the coordinates of point D

D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)

The length D'A = $\sqrt{(7)^2+(0)^2} = \sqrt{49}$ =7 units

The perimeter of polygon ABCD'

length AB + Length BC + Length CD + Length D'A

= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.

The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows

$S_{\bigtriangleup}$ = $(\frac{1}{2})|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|$

= $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

Triangle ABC we have

A(-4,-1)

B(-2,3)

C(2,2)

$S_{{\bigtriangleup}ABC}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(3-2)+(-2)(2-(-1)) +2((-1)-3)|$

= $=(\frac{1}{2})|-4-6 -8|= 9 units^2$

Triangle ACD'

A(-4,-1)

C(2,2)

D'(3, -1)

$S_{{\bigtriangleup}ACD'}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(2+1)+(2)(-1+1) +3((-1)-2)| = \frac{21}{2}$ = 10.5 units²

Area of polygon = $S_{{\bigtriangleup}ABCD'}$ = $S_{{\bigtriangleup}ABC}$ + $S_{{\bigtriangleup}ACD'}$ = (9 + 10.5) units²

Area of polygon ABCD' = 19.5 units².

7 0

2 years ago

Quadrilateral WILDWILDW, I, L, D is inscribed in circle OOO. \overline{WI} WI start overline, W, I, end overline is a diameter o

VARVARA [1.3K]

Question not well presented and diagram is missing

Quadrilateral WILD is inscribed in circle O.

WI is a diameter of circle O.

What is the measure of angle D?

See attached for diagram

Answer:

Step-by-step explanation:

Summation of opposite angles of a quadrilateral inscribed in a circle is 180°, given that the vertices are on the circle.

Given

<WIL = 45°

<ILD = 109°

In the attached;

<WIL + <WDL = 180° (Opposite angle of quadrilateral)

Substitute 45° for <WIL in the above expression

45° + <WDL = 180° ---- Collect like terms

<WDL = 180° - 45°

<WDL = 135°

Hence, the measure of angle D is 135° (See attached)

6 0

2 years ago

Reflection: Across the line y = -1

eduard

The A’ is your reflected shape. The red line is y=-1, or your line right reflection. To solve, I counted the distance from the red line on the original triangle, and then the same on the opposite side of the red line, and counted the same distance, plotting the points at the final spot.

8 0

2 years ago