Given that AD and BC bisect each other at E, prove that ⦟B ≅⦟D.

What is the solution to this inequality x-7<1

Nina [5.8K]

X−7<1

Add 7 to both sides.

x−7+7<1+7

x<8

To plot this on a number line, put a circle around 8 and a line going to the left with an arrow at the end.

8 0

3 years ago

Read 2 more answers

The equation which represents x, the width of the border is 2 left-bracket (12 + 2 x) + (15 + 2 x) right-bracket = 74. What is t

bulgar [2K]

Answer:

2 1/2

Step-by-step explanation:

If you do the step by step calculation:

Step 1: Simplify both sides of the equation.

2(12+2x+15+2x)=74

(2)(12)+(2)(2x)+(2)(15)+(2)(2x)=74(Distribute)

24+4x+30+4x=74

(4x+4x)+(24+30)=74(Combine Like Terms)

8x+54=74

Step 2: Subtract 54 from both sides.

8x+54−54=74−54

8x=20

Step 3: Divide both sides by 8.

8x divide by 8 and that is 20 divided by 8

leaves you with the answer 5/2.

If you then turn the improper fraction into a mixed fraction you get 2 and one half.

Answer= 2 1/2 or "a" DEPENDS ON WHAT QUIZ YOU HAVE

Peace!

4 0

3 years ago

Suppose that c varies jointly with d and the square of g, and c = 30 when d = 15 and g = 2.

maksim [4K]

Answer:

$d = \dfrac{3}{16}$

Step-by-step explanation:

c varies jointly with d and the square of g

$c\propto dg^2$

$c=kdg^2$

where, k is constant of proportionality.

Put the given value c = 30 when d = 15 and g = 2 and find out k

$30=k\cdot 15\cdot 2^2$

$k=\dfrac{1}{2}$

$c=\dfrac{1}{2}kg^2$

If c = 6 and g = 8 then d = ?

$6=\dfrac{1}{2}\cdot d\cdot 8^2$

$d=\dfrac{6\cdot 2}{8^2}$

$d=\dfrac{3}{16}$

Hence, The value of d is $\dfrac{3}{16}$

5 0

3 years ago

What is the constant and missing variable of these problems

Studentka2010 [4]

Answer:

14

Step-by-step explanation:

7 0

3 years ago

A 10-foot ladder is to be placed Against the side of the building. The base of the latter must be placed in an angle of 72° With

12345 [234]

Answer:

Distance of foot of ladder from building: 37.2 inches

Distance of top of ladder from building's base: 114 inches

Step-by-step explanation:

Please refer to the figure attached in the answer area.

A right angled triangle $\triangle ABC$ is formed by the ladder with the building where hypotenuse is the length of ladder.

Hypotenuse, AC = 10 foot

Also, we are given that angle made by the base of ladder with the ground is $72^\circ$ .

We have to find AB and BC.

$\angle BAC = 72^\circ$

Using trigonometric functions:

$cos \theta= \dfrac{Base}{Hypotenuse}\\cos 72^\circ= \dfrac{AB}{AC}\\\Rightarrow 0.309 = \dfrac{AB}{AC}\\\Rightarrow AB = 10 \times 0.309\\$\approx$ 3.1 foot\\$\approx$ 37.2 inches$

$sin \theta = \dfrac{Perpendicular}{Hypotenuse}\\\Rightarrow sin72^\circ = \dfrac{BC}{AC}\\\Rightarrow 0.95 = \dfrac{BC}{AC}\\\Rightarrow AB = 10 \times 0.95\\$\approx$ 9.5 foot\\$\approx$ 114 inches$

Distance of foot of ladder from building: 37.2 inches

Distance of top of ladder from building's base: 114 inches

8 0

3 years ago