All categories

4 years ago

Please help needed!!! Help me!!!

Mathematics

1 answer:

Hitman42 [59]4 years ago

5 0

a = 85°

Step-by-step explanation:

<em>Method 1</em>

5x = 2x + 57° (An exterior angle of a triangle is equal to the sum of its two interior opposite angles)

3x = 57°

x = 57/3 = 19°

a = 180 - 57 - 2x

a = 180 - 57 -2(19)

a = 85°

<em>Method 2</em>

2x + a + 57° = 180°

2x + a = 123° ==> 1

5x + a = 180°

x = (180° - a)/5 ==> 2

By substitution

2((180° - a)/5) + a = 123°

(360° - 2a)/5 + a = 123°

(360° - 2a)/5 = 123° - a

360° - 2a = 5(123° - a)

360° - 2a = 615° - 5a

-2a + 5a = 615° - 360°

3a = 255°

a = 85°

You might be interested in

5 plus 2 minus 6 times 4

nevsk [136]

Answer:

714

Step-by-step explanation:

5+2=7

7-6=1

1x4=4

3 0

4 years ago

Read 2 more answers

HELP I AM TIMED!!!!!!!!!!!!

ExtremeBDS [4]

Answer:

It is the last one 6a-4b+11

Explanation:

Hope this helps if u need clarification plz ask:)

5 0

3 years ago

2/5 into a whole number

Advocard [28]

Answer:

2/5 its a quotient just 2 divide by 5 equals to 0.4

6 0

3 years ago

Read 2 more answers

A circle has a radius of 9 inches. The Radius is multiplied by 2/3 to form a second circle. How is the ratio of the areas relate

liraira [26]

Answer:

$\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81}{36} = (\frac{r_{1} }{r_{2}}) ^{2}$

The above expression shows that ratios of the areas of the circles are equal to the square of the ratio of their radii.

Step-by-step explanation:

Radius of first circle $(r_{1})$ = 9 inches

Area of first circle = $\pi r_{1} ^{2}$

Area of first circle = 9 × 9 × π = 81 π

Now, since the radius is multiplied by 2/3 for from a new circle.

∴ Radius of the second circle = $9 \times \frac{2}{3} = 6\ inches$

Area of second circle = $\pi r_{2} ^{2}$

Area of second circle = 6 × 6 × π = 36 π

Now,

$\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81\pi }{36\pi }$

$\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81}{36} = (\frac{9}{6}) ^{2} = (\frac{r_{1} }{r_{2}}) ^{2}$

∵ $(r_{1})$ = 9 inches and $(r_{2})$ = 6 inches

The above expression shows that ratios of the areas of the circles are equal to the square of the ratio of their radii. i.e., $\frac {radius\ of\ first\ circle)^{2} }{(radius\ of\ second\ circle)^{2} } = \frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)}$

8 0

3 years ago

What is the area of a triangle with a base of 9 inches and a height of 20 inches?

andrew11 [14]

Answer:

A=90

Step-by-step explanation:

The area of a triangle with a base of 9 inches and a height of 20 inches is 90 inches.

Formula: A=hbB/2

A = hbB/2 = 20·9/2 = 90

6 0

3 years ago

Read 2 more answers