Area of the parallelogram?

[45-(6+3)]*7=<br> ???????

padilas [110]

Answer:

252

Step-by-step explanation:

[45-9]*7

36*7

252

3 0

3 years ago

What is the solution to this inequality -13x> - 39

s344n2d4d5 [400]

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable.

Inequality Form: x < 3

Interval Notation:

( − ∞ , 3 )

3 0

3 years ago

Read 2 more answers

The yearly cost y for driving a certain car is $7,800 for 10,000 miles

Licemer1 [7]

Answer:

y = 0.36x + 4200.00

15100.00 miles.

Step-by-step explanation:

If the yearly cost for driving x miles is given by y dollars then the two ordered pairs to find the linear relation between x and y are (10000,7800) and (20000,11400).

So, the equation will be

$\frac{y - 11400}{11400 - 7800} = \frac{x - 20000}{20000 - 10000}$

⇒ $\frac{y - 11400}{3600} = \frac{x - 20000}{10000}$

⇒ $y - 11400 = \frac{9}{25}(x - 20000)$

⇒ $y - 11400 = \frac{9}{25}x - 7200$

⇒ $y = \frac{9}{25}x + 4200$

⇒ y = 0.36x + 4200.00 ........... (1) (Answer)

Now, for y = $9636.00, from equation (1),

9636 = 0.36x + 4200

⇒ 0.36x = 5436

⇒ x = 15100.00 miles. (Answer)

6 0

2 years ago

If D is midpoint of side BC of triangle ABC. P and Q are points lying respectively on side AB and AC such that DP is parallel to

LUCKY_DIMON [66]

See proof below

Step-by-step explanation:

Assume triangle ABC to have vertices at;

A(2,-1), B(2,-7) and C(6,-7)

D is midpoint of BC, thus D is at (4,-7)

The P and Q, are lying on side AB and AC, hence assume P is at (2,-4) and Q is at (4,-4) such at DP is parallel to QA

Plot the points on a graph tool and join the points to view the sketch.

To prove area of triangle CPQ is 1/4 area of ABC will be;

Find area ABC and CPQ then compare the areas.

Apply the distance formula to find the length of sides of the triangles then find the areas.

The distance formula is;

$d=\sqrt{x_2-x_1)^2+(y_2-y_1)^2}$

Length of side AB from the sketch is;

$AB=\sqrt{(2-2)^2+(-7--1)^2} =\sqrt{-6^2} =\sqrt{36} =6units$

Length of side BC will be;

$BC=\sqrt{(-7--7)^2+(6-2)^2} =\sqrt{4^2} =\sqrt{16} =4units$

Thus area of triangle ABC will be;

1/2 *base length*height ------because it is a right-triangle

1/2*4*6=12 square units

Find the lengths of all sides of triangle CPQ

Length of side PQ is half that of side BC thus PQ=2 units

Length of side PC is;

$PC=\sqrt{(6-2)^2+(-7--4)^2} =\sqrt{4^2+-3^2} =\sqrt{16+9} =\sqrt{25} =5units$

Length of side QC will be;

$QC=\sqrt{(6-4)^2+(-7--4)^2} =\sqrt{2^2+-3^3} =\sqrt{4+9} =\sqrt{13}$

QC= √13 = 3.6 units

Find area of triangle CPQ given all sides by applying the Heron's formula for area of triangle which is;

A=√s(s-a)(s-b)(s-c) where;

A=area of the triangle

s= half the perimeter of the triangle

a=side PQ = 2 units

b=side PC = 5 units

c= side QC = 3.6 units

Finding the perimeter of triangle CPQ will be;

P=sum of all sides

P=2+5+3.6 =10.6 units

s=10.6/2 = 5.3

Area of the triangle CPQ will be;

$A=\sqrt{5.3(5.3-2)(5.3-5)(5.3-3.6)} \\A=\sqrt{5.3(3.3)(0.3)(1.7)} \\A=\sqrt{8.9} =2.98$

A=3.0 (1 decimal place)

Compare the areas;

Area of triangle ABC=12 square units

Area of triangle CPQ = 3 square units

Area of triangle CPQ / Area of triangle ABC = 3/12 =1/4

Thus you have proved that area of triangle CPQ is 1/4 th area of triangle ABC because 1/4 *12 =3

Learn More

Area of a triangle ;brainly.com/question/14869984

The Heron's formula : brainly.com/question/10713495

Keywords: midpoint, triangle, sides, parallel, prove , area, equal

#LearnwithBrainly

6 0

2 years ago

Anyone know this answer?

Alexeev081 [22]

Answer:

try -8/-5 -10

Step-by-step explanation:

7 0

2 years ago