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Nat2105 [25]
3 years ago
9

Solve the equation for y 4x+2y=38

Mathematics
1 answer:
GenaCL600 [577]3 years ago
3 0

Answer

when mommy a daddy love each other very much they make love to each other

Step-by-step explanation:

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sleet_krkn [62]
Answer=
426cm squared
3 0
3 years ago
How would the following triangle be classified?
Anna [14]

Answer:

hope this helpful

Step-by-step explanation:

isosceles right

because A light ray incident at 90° at the first face emerges at the same angle. The diagram shows five isosceles right-angled prisms. A light ray incident at 90° at the first face emerges at the same angle with the normal from the last face.

8 0
3 years ago
Solve the inequality and express in interval notation and graph its solution on a number line
vfiekz [6]

Answer:

x ∈ (-∞, 3) U (6, ∞).

Step-by-step explanation:

x^2> 9x - 18

x^2 - 9x + 18> 0

We use factorization and optain

(x-6)(x-3)> 0

Then, we have two critical points: x=3 and x=6. Now:

(i) for x < 3 we have that x-6 <0 and x-3 <0. Then (x-6)(x-3)  > 0.

(ii) for 3 < x < 6 we have that x -6 <0 and x -3 > 0. Then (x-6)(x-3)  < 0.

(iii) for x > 6 we have that x-6 >0 and x-3 > 0. Then, (x-6)(x-3)  > 0.

conditions (i) and (iii) satisfy the inequatliy, then the solution is x ∈ (-∞, 3) U (6, ∞).

The graph is in the picture below.

7 0
3 years ago
Together, the areas of the rectangles sum to 30 square centimeters.
Juli2301 [7.4K]

Answer:

3x+2y

Step-by-step explanation:

Find the rectangles attached

Area of a rectangle = Length * Width

A = A1+A2

given

A = 30cm²

A1 = 3 * x

A1 = 3x

A2 = 2*y

A2 = 2y

Substitute

30 = 3x+2y

Hence the required equation is 3x+2y = 30

6 0
3 years ago
Line Segment DE is parallel to side BC of right triangle ABC. CD = 3, DE = 6, and EB = 4. Compute the area of quadrilateral BCDE
dybincka [34]

The area of quadrilateral BCDE = 20.4 sq. units

Let AD = x and AE = y.

Since ΔABC and ΔAED are similar right angled triangles, we have that

AC/AD = AB/AE

AC = AD + CD

= x + 3.

Also, AB = AE + EB

= y + 4

So, AC/AD = AB/AE

(x + 3)/x = (y + 4)/y

Cross-multiplying, we have

y(x + 3) = x(y + 4)

Expanding the brackets, we have

xy + 3y = xy + 4x

3y = 4x

y = 4x/3

In ΔAED, AD² + AE² = DE².

So, x² + y² = 6²

Substituting y = 4x/3 into the equation, we have

x² + y² = 6²

x² + (4x/3)² = 6²

x² + 16x²/9 = 36

(9x² + 16x²)/9 = 36

25x²/9 = 36

Multiplying both sides by 9/25, we have

x² = 36 × 9/25

Taking square root of both sides, we have

x = √(36 × 9/25)

x = 6 × 3/5

x = 18/5

x = 3.6

Since y = 4x/3,

Substituting x into the equation, we have

y = 4 × 3.6/3

y = 4.8

To find the area of quadrilateral BCDE, we subtract the area of ΔAED from area of ΔABC.

So, area of quadrilateral BCDE = area of ΔABC - area of ΔAED

area of ΔABC = 1/2 AC × AB

= 1/2 (x + 3)(y + 4)

= 1/2(3.6 + 3)(4.8 + 4)

= 1/2 × (6.6)(8.8)

= 1/2 × 58.08

= 29.04  square units

area of ΔAED = 1/2 AD × AE

= 1/2xy

= 1/2 × 3.6 × 4.8

= 1/2 × 17.28

= 8.64 square units

area of quadrilateral BCDE = area of ΔABC - area of ΔAED

area of quadrilateral BCDE = 29.04 sq units - 8.64 sq units

area of quadrilateral BCDE = 20.4 sq. units

So, the area of quadrilateral BCDE = 20.4 sq. units

Learn more about area of a quadrilateral here:

brainly.com/question/19678935

4 0
2 years ago
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