Write the first 8 triangular numbers

[tex]\mathsf\blue{♧ANSWER♧}[/tex]

[tex] \mathsf \orange{ \frac{x \times y}{2}} [/tex]

[tex] \mathsf \green{ \frac{8 \times 12}{2} }[/tex]

[tex] \mathsf \orange{ \frac{96}{2} }[/tex]

[tex] \mathsf \green{48 \: square \: units}[/tex]

[tex] \huge \tt \color{pink}{A}\color{blue}{n}\color{red}{s}\color{green}{w}\color{grey}{e}\color{purple}{r }[/tex]

[tex]\qquad\qquad\qquad \large\underline{ \boxed{ \sf{✰\: Information }}}[/tex]

[tex]{ \boxed{ ✟ \: \underline{ \boxed{ \sf \: Area \: of \: kite = \frac{ d_1 \times d_2}{2} \: or \blue{ \: \frac{1}{2} \times d_1 \times d_2}}}✟ }}[/tex]

[tex]\rule{80mm}{2.5pt}[/tex]

[tex] \qquad \rm{➛Area \: of \: kite = \frac{BD×AC}{2} }[/tex]

[tex]\qquad \rm{➛Area \: of \: kite = \frac{8×12}{2} } \\ \qquad \rm{➛Area \: of \: kite = \frac{ \cancel{96}}{ \cancel2} } \\ \qquad \rm{➛Area \: of \: kite = 48 {units}^{2} }[/tex]

[tex]\rule{80mm}{2.5pt}[/tex]

★Hence area of kite with diagonals BD = 8 units and AC = 12 units is

[tex]\qquad\qquad\qquad{ \boxed{↪ \underline{ \boxed{ \sf{\: 48 {units}^{2} \green✓ }}}↩}}[/tex]

[tex]\rule{80mm}{2.5pt}[/tex]

★ More info regarding this topic ★

[tex]\rule{80mm}{2.5pt}[/tex]

Hope it helps !

Step-by-step explanation:

For question 22: Notice the congruent symbol on sides AB and CB; this shows that the triangle is isosceles, and in an isosceles triangle, the angles opposite congruent sides are congruent. Thus, angle A actually equals angle C.

In other the find B, it is know that all of the interior angles of a triangle must add to 180; thus, this can be written as: angle A + angle B + angle C = 180.

Since you already know angles A and C, subtract angle A and angle C from 180 to get angle B.

For question 23:

Notice that triangle DBC is an isosceles triangle, and that angle D and angle C are opposite of congruent sides, indicating that they are equal. Since the interior angles of a triangle must add up to 180, and you know that angle D equals angle C, you can find angle DBC by doing 180 - (2 * angle D).

Now that you have angle DBC, you can find angle CBA; since these two angles form a straight line, you know they must add to 180 degrees. Thus, 180 - angle DBC equals angle CBA.

Now, notice that triangle CBA is also isosceles, and that angles CBA and angle A are opposite congruent sides, indicating that they are also congruent. Thus, angle A equals angle CBA.

Hope this helps :)

Answer:

  period

Step-by-step explanation:

The attached graph shows a period of the function. Its amplitude is 1/4 (vertical distance from centerline to peak). Its period is 4π (horizontal distance from peak to peak). Its frequency is the reciprocal of the period, so would be 1/(4π).

The value 4π is the period of the function.

Answer:

[tex]\displaystyle 11,3[/tex]

Step-by-step explanation:

Use the Law of Cosines to find the length of the third edge:

Solving for Angles

[tex]\displaystyle \frac{a^2 + b^2 - c^2}{2ab} = cos\angle{C} \\ \frac{a^2 - b^2 + c^2}{2ac} = cos\angle{B} \\ \frac{-a^2 + b^2 + c^2}{2bc} = cos\angle{A}[/tex]

Use [tex]\displaystyle cos^{-1}[/tex]towards the end or you will throw your result off!

Solving for Edges

[tex]\displaystyle b^2 + a^2 - 2ba\:cos\angle{C} = c^2 \\ c^2 + a^2 - 2ca\:cos\angle{B} = b^2 \\ c^2 + b^2 - 2cb\:cos\angle{A} = a^2[/tex]

Take the square root of the final result or it will be thrown off!

Let us get to wourk:

[tex]\displaystyle 10^2 + 7,9^2 - 2[10][7,9]\:cos\:77,5 = a^2 \hookrightarrow 100 + 62,41 - 158\:cos\:77,5 = a^2 \hookrightarrow \sqrt{128,212541} = \sqrt{a^2}; 11,323097677... \\ \\ \boxed{11,3 \approx a}[/tex]

I am joyous to assist you at any time.

Answer:

x= 12

Step-by-step explanation:

Hope u understood it

"PLEASE MARK ME THE BRAINLIEST"

[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

As we know ~

Area of the circle is :

[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

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[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 4.4\div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 2.2 \: mm[/tex]

Now find the Area ~

[tex]\qquad \sf  \dashrightarrow \: \pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {(2.2)}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {4.84}^{} [/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2} [/tex]

・ ．━━━━━━━†━━━━━━━━━．・

[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 3.7 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 1.85 \: \: cm[/tex]

Bow, calculate the Area ~

[tex]\qquad \sf  \dashrightarrow \: \pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times (1.85) {}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 3.4225 {}^{} [/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2} [/tex]

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[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times (8.3) {}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 68.89[/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx216.31 \: \: cm {}^{2} [/tex]

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[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 5.8 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 2.9 \: \: yd[/tex]

now, let's calculate area ~

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {(2.9)}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 8.41 [/tex]

[tex] \qquad \sf  \dashrightarrow \:area \approx26.41 \: \: yd {}^{2} [/tex]

・ ．━━━━━━━†━━━━━━━━━．・

[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 1 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 0.5 \: \: yd[/tex]

Now, let's calculate area ~

[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times (0.5) {}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 0.25[/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx0.785 \: \: yd {}^{2} [/tex]

・ ．━━━━━━━†━━━━━━━━━．・

[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {(8)}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 64[/tex]

[tex]\qquad \sf  \dashrightarrow \:area = 200.96 \: \: yd {}^{2} [/tex]

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Step-by-step explanation:

1,186,200,000,000 in standard form

Answer:

2.7

Step-by-step explanation:

x=4

also uhh where does the y go-

Answer:

18 mi^2

...

...

...

...

...

Answer:

  42 square miles

Step-by-step explanation:

The area of a composite shape can be found by decomposing it into shapes whose area formulas you know. Here, the shape can be decomposed into a triangle and two rectangles. The two rectangles can be formed several ways.

The area of the triangle at the top of the figure is ...

  A = 1/2bh

  A = 1/2(6 mi)(3 mi) = 9 mi²

The area of the overall rectangle (including the space at lower right) is ...

  A = LW

  A = (9 mi)(5 mi) = 45 mi²

The area of the rectangle at lower right that is not included in the shaded area is ...

  A = LW

  A = (4 mi)(3 mi) = 12 mi²

The total area of the figure is the sum of the triangle and overall rectangle, less the excluded area at lower right.

  total area = 9 mi² +45 mi² -12 mi² = 42 mi²

The area of the figure is 42 square miles.

_____

Additional comment

Other ways to divide the figure into two rectangles are ...

Answer:

16. j = 105 °

17. Q = 68 °

18. s = 97 °

19. t = 70 °

20. r = 125 °

Step-by-step explanation:

16. Name the angle that is supplementary to angle J as K, and label the angle that is next to K as G

Angle G = 105, because it is corresponding to the 105 angle.

Since G is supplementary to K, angle K = 75.

So, J = 105

17. Name the angle that is supplementary to angle Q as R

Angle R = 112, because it is corresponding to the 112 angle.

Since R is supplementary to Q, angle Q = 68.

18. Name the angle that is supplementary to the 83 angle as L
Angle L = 97, since angle L and the 83 angle are supplementary angles

-> Angle S = 97, since S and L are alternate angles

19. Angle T = 70, since angle T and the 70 angle are corresponding angles.

20. Name the angle that are alternate angles with angle R as Q

Angle Q = 125, because it is corresponding to the 125 angle.

Since angle Q and angle R are alternate angles, angle R = 125.

(Note : alternate angles and corresponding angles are equal to each other, supplementary angles add up to 180.)

Answer:

2929 1/3 [tex]\pi[/tex] in^3

Step-by-step explanation:

The equation for the volume of a sphere is [tex]\frac{4}{3} \pi r^3[/tex].

Insert 13 for r.  Then, simplify.

Answer:

f(-6)= -41 923

Step-by-step explanation:

f(-6)= -( -6)⁶ +5(-6)⁴+8(-6)³+2(-6) -7 you substitute x with -6 the you use a calculator which will then give you - 41923

Answer:

31

Step-by-step explanation:

[tex]4^{2}[/tex]=16

16+15=31

Answer:

5.6

Step-by-step explanation:

4/5 = .8

so ? ÷ 7 = .8|
7 x .8 = 5.6
x= 5.6

x = 5.6

hope it helps...!!!!

Answer:

As per Provided Information

Height of cone h is 3.25 cm

Diameter of cone is 2.2 cm

Radius = 2.2/2

Radius = 1.1 cm

We have been asked to determine the volume of the cone .

Using Formulae

[tex] \boxed{\bf \:Volume_{(Cone)} = \cfrac{1}{3} \pi {r}^{2}h}[/tex]

Substituting the given value and let's solve it

[tex]\sf\longrightarrow \: Volume_{(Cone)} = \cfrac{1}{3} \times 3.14 \times {1.1}^{2} \times 3.25 \\ \\ \\ \sf\longrightarrow \: Volume_{(Cone)} = \cfrac{1}{3} \times 3.14 \times 1.21 \times 3.25 \\ \\ \\ \sf\longrightarrow \: Volume_{(Cone)} = \cfrac{1}{3} \times 3.14 \times 3.9325 \\ \\ \\ \sf\longrightarrow \: Volume_{(Cone)} = \cfrac{1}{3} \times 12.34805 \\ \\ \\ \sf\longrightarrow \: Volume_{(Cone)} = \cfrac{12.34805}{3} \\ \\ \\ \sf\longrightarrow \: Volume_{(Cone)} = 4.11 {cm}^{3} [/tex]

Therefore,

5 quarts I think 4 cups per quart has 5 quarts and 2 cups I think

Answer:

1 quart = 4 cups

8 × 4 = 32

32 - 6 = 26

26 - 18 = 8

he has 2 quarts left

Answer:

  0.50 seconds

Step-by-step explanation:

The time it takes for an echo to return is the time it takes for the sound to make the round trip between the source and the reflecting object.

The boulder is 80 m away from the source, so the round-trip distance is 2×80 m = 160 m. At a speed of 320 m/s, it takes sound ...

  (160 m)/(320 m/s) = 0.50 s

to make the round-trip.

It will take 0.50 seconds to hear the echo.

The time it will take to hear the echo, when the speed of a sound wave is 320 m/s is 0.50 seconds.

What is the speed of sound?

Speed of sound is the rate by which the sound wave travels. The speed of sound depends on the medium by which the sound wave is travelling.

The speed of the wave can be found out using the following formula.

[tex]s=\dfrac{d}{t}[/tex]

Here, (d) is the distance travelled by the wave and (t) is time taken by the wave to cover that distance.

A boy yells in a narrow valley. The sound reflects off of a boulder that is 80 m away.

For the echo, distance travelled will be doubled as it come back.Thus, the distance travelled by sound wave is,

[tex]d=2\times80\\d=160\rm\; m[/tex]

The speed of sound is 320 m/s. Put these value in the above formula as,

[tex]320=\dfrac{160}{t}\\t=0.5\rm\; s[/tex]

Thus, the time it will take to hear the echo, when the speed of a sound wave is 320 m/s is 0.50 seconds.

Learn more about the speed of sound here;

https://brainly.com/question/95976

[tex]\qquad\qquad\huge\underline{\boxed{\sf Answer}}[/tex]

The sign of a will be negative ~

let the value of be be ' -x ' , -x is a negative number (b = -x < 0)

[tex]\qquad \sf  \dashrightarrow \: a \times \frac{ - b}{b} [/tex]

[tex]\qquad \sf  \dashrightarrow \: a \times \frac{ - ( - x)}{ - x} [/tex]

[tex]\qquad \sf  \dashrightarrow \: a \times \frac{ x}{ - x} [/tex]

[tex]\qquad \sf  \dashrightarrow \: a \times \frac{ - x}{ x} [/tex]

[tex]\qquad \sf  \dashrightarrow \:a \times - 1[/tex]

[tex] \qquad \sf  \dashrightarrow \:- a[/tex]

Therefore, the term a, has a negative sign unless it's value is substituted, I.e -a = 0 ( having no sign )

Answer:

56

Step-by-step explanation:

Answer:

6×6=36

36 is the perfect square

Step-by-step explanation:

x+134=180(straight angle)

x=46

Answer:

[tex]V=\pi r^2h=\pi *6^2*19[/tex] ≈ [tex]2148.8[/tex]

Let me know if you found this helpful! It means a lot!

The calculated areas of both are the same instead of using the different formulas that are 43. 5 squared unit.

The area of the polygon is the product of half of the apothem of the triangle and perimeter.

The area of the polygon = [tex]\frac{1}{2} \times h \times p[/tex]

The side of the triangle is 10 units.

Apothem of an equilateral triangle

[tex]s = \frac{\sqrt{3} a}{6}[/tex]

s= [tex]\sqrt{3} \times 10/6[/tex]

s = 2.88

Here, a is the side of the triangle.

Thus the apothem of the triangle is 2.9 units.

The perimeter of the equilateral triangle is equal to the three times the sides.

Perimeter of the equilateral triangle ,

P = 3(10)

P = 30

Therefore, the perimeter of the equilateral triangle is 30 units.

The area of the equilateral triangle-

The area of the polygon = [tex]\frac{1}{2} \times h \times p[/tex]

A = [tex]1/2 \times 2.9 \times 30[/tex]

A = 43.5

Thus the area of the equilateral triangle is 43.5 squared units.

Therefore, The calculated areas of both are the same instead of using the different formulas.

Learn more about the apothem;

brainly.com/question/10580427

Answer:

(A) 2.9

(C)30

(A)1/2(2.9)(30)

(D)The same despite using different formulas

Step-by-step explanation:

I did it

Answer:

Step-by-step explanation:

you would multiply the fractions

then you would add the hole number

Answer:

2 14/15

Step-by-step explanation:

Answer:

Pretty sure it's 2:30 P.M.

Step-by-step explanation:

If her track practice started at 12:00 P.M., and and ran relays for one hour, it would already be 1:00 P.M.

Then she practiced the long jump for an hour and thirty minutes. Add another hour to that and it would be 2:00 P.M. Add thirty minutes, and it would be 2:30 P.M.

Hopefully that makes sense to you

Answer:

5π/6

Step-by-step explanation:

Given :

Solving :

Answer:

5π Over 6

Step-by-step explanation:

You can write  7π6 as (π+π6)Thus we can clearly see the angle falls in the third quadrant. And the cosine value in third quadrant is always negative. Hence, cos(π+π6)=−cos(π6)coming back to the question cos−1[cos(7π6)]=cos−1[−cos(π6)]=π−cos−1[cos(π6)]=π−π6=5π6

Answer:

d = st, d = 100 x 3

Step-by-step explanation:

distance = speed x time

distance = 100 x 3, therefore, distance = 300, but if you just want the equation, it is d = 100 x 3 or d = st

Step-by-step explanation:

d = 100 x 3.

d= 100n

where n = time the journey took.