Which graph represents the solution set for the system x + y> 5 and -3x + 2y< -2.

Graph ABCD if A(1, 4), B(6, 6), C(4, 1), and D(−1, −1). Is ABCD a rhombus? Show how you know.

stich3 [128]

C4,1 because I solved it and it showed me

6 0

3 years ago

Solve by substitution <br> x - 2y = 10<br> 1/2x - 2y = 4

mars1129 [50]

Answer:

ypu can do this by using this please vote

5 0

3 years ago

Amelia measured the elementary school and made a scale drawing. The scale of the drawing was 11 mm, 4 m. If the schoolyard is 48

cestrela7 [59]

Answer:

176

Step-by-step explanation:

every 11 mm in the drawing equals 4m.

every 1 mm in the drawing equals 4/11 m

Actual size of school yard = 484*4/11 = 176m

3 0

2 years ago

The diagram shows a regular octagon ABCDEFGH. Each side of the octagon has length 10cm. Find the area of the shaded region ACDEH

Zolol [24]

The area of the shaded region /ACDEH/ is 325.64cm²

Step 1 - Collect all the facts

First, let's examine all that we know.

We know that the octagon is regular which means all sides are equal.
since all sides are equal, then all sides are equal 10cm.
if all sides are equal then all angles within it are equal.
since the total angle in an octagon is 1080°, the sum of each angle within the octagon is 135°.

Please note that the shaded region comprises a rectangle /ADEH/ and a scalene triangle /ACD/.

So to get the area of the entire region, we have to solve for the area of the Scalene Triangle /ACD/ and add that to the area of the rectangle /ADEH/

Step 2 - Solving for /ACD/

The formula for the area of a Scalene Triangle is given as:

A = $\sqrt{S(S-a)(S-b)(S-c) square units}$

This formula assumes that we have all the sides. But we don't yet.

However, we know the side /CD/ is 10cm. Recall that side /CD/ is one of the sides of the octagon ABCDEFGH.

This is not enough. To get sides /AC/ and /AD/ of Δ ACD, we have to turn to another triangle - Triangle ABC. Fortunately, ΔABC is an Isosceles triangle.

Step 3 - Solving for side AC.

Since all the angles in the octagon are equal, ∠ABC = 135°.

Recall that the total angle in a triangle is 180°. Since Δ ABC is an Isosceles triangle, sides /AB/ and /BC/ are equal.

Recall that the Base angles of an isosceles triangle is always equal. That is ∠BCA and ∠BAC are equal. To get that we say:

180° - 135° = 45° [This is the sum total of ∠BCA and ∠BAC. Each angle therefore equals

45°/2 = 22.5°

Now that we know all the angles of Δ ABC and two sides /AB/ and /BC/, let's try to solve for /AC/ which is one of the sides of Δ ACD.

According to the Sine rule,

$\frac{Sin 135}{/AC/}$ = $\frac{Sin 22.5}{/AB/}$ = $\frac{Sin 22.5}{/BC/}$

Since we know side /BC/, let's go with the first two parts of the equation.

That gives us $\frac{0.7071}{/AC/} = \frac{0.3827}{10}$

Cross multiplying the above, we get

/AC/ = $\frac{7.0711}{0.3827}$

Side /AC/ = 18.48cm.

Returning to our Scalene Triangle, we now have /AC/ and /CD/.

To get /AD/ we can also use the Sine rule since we can now derive the angles in Δ ABC.

From the Octagon the total angle inside /HAB/ is 135°. We know that ∠HAB comprises ∠CAB which is 22.5°, ∠HAD which is 90°. Therefore, ∠DAC = 135° - (22.5+90)

∠DAC = 22.5°

Using the same deductive principle, we can obtain all the other angles within Δ ACD, with ∠CDA = 45° and ∠112.5°.

Now that we have two sides of ΔACD and all its angles, let's solve for side /AD/ using the Sine rule.

$\frac{Sin 112.5}{/AD/}$ = $\frac{Sin 45}{18.48}$

Cross multiplying we have:

/AD/ = $\frac{17.0733}{0.7071}$

Therefore, /AD/ = 24.15cm.

Step 4 - Solving for Area of ΔACD

Now that we have all the sides of ΔACD, let's solve for its area.

Recall that the area of a Scalene Triangle using Heron's formula is given as

A = $\sqrt{S(S-a)(S-b)(S-c) square units}$

Where S is the semi-perimeter given as

S= (/AC/ + /CD/ + /DA/)/2

We are using this formula because we don't have the height for ΔACD but we have all the sides.

Step 5 - Solving for Semi Perimeter

S = (18.48 + 10 + 24.15)/2

S = 26.32

Therefore, Area = $\sqrt{26.32(26.32-18.48)(26.32-10)(26.32-24.15)}$

A = $\sqrt{26.32 * 7.84*16.32 * 2.17)}$

A = $\sqrt{7,307.72}$ Square cm.

A of ΔACD = 85.49cm²

Recall that the shape consists of the rectangle /ADEH/.

The A of a rectangle is L x B

A of /ADEH/ = 240.15cm²

Step 6 - Solving for total Area of the shaded region of the Octagon

The total area of the Shaded region /ACDEH/, therefore, is 240.15 + 85.49

= 325.64cm²

See the link below for more about Octagons:
brainly.com/question/4515567

8 0

2 years ago

PLEASE HELP ASAP!!!

EleoNora [17]

<span>Part 1:

$(x+1)^2-3(x+2) \\ \\ =x^2+2x+1-3x-6 \\ \\ =x^2-x-5$

option b.

Part 2:
$(5x^3y^{-3})(-6x^4y)\\ \\=5(-6)x^{3+4}y^{-3+1}\\ \\=-30x^7y^{-2}\\ \\=\frac{-30x^7}{y^2}$

option d

Part 3:
$\frac{8+x}{2} - \frac{x+3}{3} =4$

Multiply through by the LCM of 2 and 3 (i.e. 6)
$6\left(\frac{8+x}{2}\right) - 6\left(\frac{x+3}{3}\right) =6(4) \\ \\ 3(8+x)-2(x+3)=24 \\ \\ 24+3x-2x-6=24 \\ \\ x+18=24 \\ \\ x=24-18=6$

option b

Part 4
x^2-3x-8=x+4

$x^2-3x-8=x+4 \\ \\ x^2-3x-x-8-4=0 \\ \\ x^2-4x-12=0 \\ \\ (x-6)(x+2)=0 \\ \\ x=6 \ or \ -2$

option d

Part 5
Joseph is solving the equation x^2+8x-4=0 using the technique completing the square.
x^2+8x-4=0
x^2+8x+?=4+?

In completeing the square method, you divide the coeficient of x by 2, square the result and add to both sides of the equation.

Therefore, he should add 16 to both sides because (8/2)^2=16.

option d.
</span>

5 0

3 years ago