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3 years ago

14

Which lines have a y intercept of 5

1 answer:

stiks02 [169]3 years ago

8 0

Y=2x+5 could be used for this since it has a y-int of 5

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PLEASE HELP WILL MARK BRAINLIEST

LekaFEV [45]

Step-by-step explanation:

In trapezoid ABCD, it is given that:

AE = BE & DF = CF

Hence, points E and F are the mid points of the sides AB and CD respectively.

We know that:

In a trapezoid length of the line segment joining the mid points of non parallel sides is equal to one half of the sum of lengths of the parallel sides.

$\therefore \: 4x = \frac{1}{2} \{(x + 5) + 44 \} \\ \\ \therefore \: 4x = \frac{1}{2} \{x + 5 + 44 \} \\ \\ \therefore \: 4x \times 2= \{x + 49\} \\ \\ \therefore \: 8x = x + 49 \\ \\ \therefore \: 8x - x = 49 \\ \\ \therefore \: 8x - x = 49 \\ \\ \therefore \:7x = 49 \\ \\ \therefore \:x = \frac{49}{7} \\ \\ \therefore \:x = 7\\ \\ \therefore \:4x = 4\times 7.. (\because EF = 4x)\\ \\ \: \: \: \: \: \huge \red{ \boxed{\therefore \:EF = 28}}$

3 0

4 years ago

What are the numerical measures of the angles?

vredina [299]

ANSWER

The sum of m∠DEA and m∠AEB is 180°.

Reason: The sum of angles on a straight line is always 180°.

$(x + 25)+(9x + 10)=180 \degree$

Group like terms to get,

$x+9x=180-25-10$

Simplify, and get,

$10x = 145$

Divide both sides by 10 to get,

$x = 14.5$

We substitute this value to get the required angles.

$m \: ∠DEA=x+25$

$m \: ∠DEA = 14.5 + 25 = 39.5 \degree$

$m\:∠AEB=9x+ 10$

$m\:∠AEB = 9(14.5) + 10$

$m\: ∠AEB=130.5 + 10 = 140.5 \degree$

7 0

3 years ago

Read 2 more answers

What’s 2.4 x 10to the power of 12

max2010maxim [7]

2.4 x 10^12 = 2,400,000,000,000

Hope this helped!!

6 0

3 years ago

Read 2 more answers

4=y divided by 3. what is y

Marrrta [24]

Answer:

y = 9

Step-by-step explanation:

1) Multiply both sides by 3.

3 × 3 = y

2) Simplfiy 3 × 3 to 9.

9 = y

3) Switch sides.

9

6 0

3 years ago

Read 2 more answers

PLEASE ILL MAKE BRAINIEST!!!

expeople1 [14]

Answer:

The area of rectangle BEFD is 180 square units.

Step-by-step explanation:

After checking the figure given, we have the following information:

$AD = 15$ , $AB = 12$ , $BC = 15$ , $CD = 12$

By Pythagorean Theorem, we determine the length of the line segment BD:

$BD = \sqrt{AB^{2}+AD^{2}}$ (1)

$BD = \sqrt{12^{2}+15^{2}}$

$BD = 3\sqrt{41}$

In addition, we know the following characteristics of the rectangle BEFD:

$BD = EF$ , $EF = EC + CF$ , $BE = FD$ (2), (3), (4)

By Pythagorean Theorem:

$BC^{2} = BE^{2}+EC^{2}$ (5)

$CD^{2} = CF^{2}+DF^{2}$ (6)

By (3), (4), (5) and (6):

$BC^{2} = BE^{2} + EC^{2}$ (7)

$CD^{2} = (EF-EC)^{2} + BE^{2}$ (8)

By (7) in (8):

$CD^{2} = (EF-EC)^{2}+ (BC^{2}-EC^{2})$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC + EC^{2}+BC^{2}-EC^{2}$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC +BC^{2}$

Then, we clear $EC$ :

$2\cdot EF\cdot EC = EF^{2} + BC^{2} - CD^{2}$

$EC = \frac{EF^{2}+BC^{2}-CD^{2}}{2\cdot EF}$

If we know that $EF = 3\sqrt{41}$ , $BC = 15$ and $CD = 12$ , then the length of the segment $EC$ is:

$EC = \frac{75\sqrt{41}}{41}$

And the length of the line segment $CF$ is:

$CF = EF - EC$

$CF = 3\sqrt{41}-\frac{75\sqrt{41}}{41}$

$CF = \frac{48\sqrt{41}}{41}$

And the length of the line segment $DF$ is determined by Pythagorean Theorem:

$FD = \sqrt{CD^{2}-CF^{2}}$

$FD = \sqrt{12^{2}-\left(\frac{48\sqrt{41}}{41} \right)^{2}}$

$FD = \frac{60\sqrt{41}}{41}$

And the area of the rectangle is determined by the following formula:

$A = FD\cdot EF$

$A = \left(\frac{60\sqrt{41}}{41} \right)\cdot (3\sqrt{41})$

$A = 180$

The area of rectangle BEFD is 180 square units.

4 0

3 years ago