Which must be true ?

The radius of a circle is 2.8 cm. Find the circumference to the nearest tenth.

Neporo4naja [7]

Answer: 17.59 cm

Step-by-step explanation: C=2πr=2·π·2.8=17.59292 cm

6 0

2 years ago

Read 2 more answers

How do you find the angles in a triangle made up of two isosceles triangles? I know that one is 80 degrees but i don't know how,

Stolb23 [73]

I think i can do it better with a photo of the math problem, but asap u post it, tell me!! I'd love to help!

6 0

3 years ago

Is 1/8 grader than or less than 2/5

Oduvanchick [21]

$\large\text{Hey there!}$

$\mathsf{Is\ \dfrac{1}{8}\ greater\ than\ or\ less\ than\ \dfrac{2}{5}\ ?}$

$\large\text{Personally, if you convert your fractions into decimals it can be easier to}\\\large\text{to answer your question. Or you could even match it with the LCD}\\\large\text{if it is easier for you!}$

$\mathsf{\dfrac{1}{8}=0.125} \\\\\mathsf{\dfrac{2}{5}=0.40}$

$\large\text{The\ LCD (Lowest Common Denominator) is 40}$

$\mathsf{Match\ your\ fractions\ with\ your\ LCD}\\\\\boxed{\mathsf{\dfrac{1}{8}}}\boxed{\mathsf{\dfrac{2}{5}}}\\\boxed{\mathsf{\dfrac{5}{40}}}\boxed{\mathsf{\dfrac{16}{40}}}$

$\large\text{We compared }\mathsf{\dfrac{1}{8}\ \&\ \dfrac{5}{40}.}\large\text{ We also compared }\mathsf{\dfrac{2}{5} \ \& \ \dfrac{16}{40}}$

$\mathsf{Both\ \dfrac{2}{5} \& \ \dfrac{16}{40}\ are\ LESS/ SMALLER\ than\ \dfrac{1}{8} \ & \ \dfrac{5}{40}}$

$\boxed{\boxed{\large\text{Thus your answer is: \boxed{\mathsf{\dfrac{1}{8} \ is\ GREATER\ than\ \dfrac{2}{5}}}}}}}\huge\checkmark$

$\large\text{Good luck on your assignment and enjoy your day!}$

~ $\large\dag\dfrac{\frak{LoveYourselfFirst}}{:)}$

3 0

3 years ago

Drag Each expression to show whether it is equivalent to 6.7-(-2.5), 6.7+(-2.5), or neither.

iogann1982 [59]

Answer:

Step-by-step explanation:

6.7 – (-2.5)

2.5+6.7

6.7 + (-2.5)

-2.5+6.7

6.7-2.5

neither

-2.5-6.7

2.5-6.7

-6.7+2.5

4 0

3 years ago

|x-3|<2 |4x+1|>0 |x-1|<5 Ayudaaaaaaa pleaseeeee

Dafna11 [192]

Recuerda que

• |x| = x si x ≥ 0

• |x| = -x si x < 0

Necesitas considerar dos casos:

• si x - 3 ≥ 0,

|x - 3| < 1 ⇒ x - 3 < 1 ⇒ x < 4

• si x - 3 < 0,

|x - 3| < 1 ⇒ -(x - 3) = 3 - x < 1 ⇒ -x < -2 ⇒ x > 2

Entonces la solución consta de todos los números reales x tales que x > 2 y x < 4, o simplemente 2 < x < 4.

El método para resolver las otras desigualdades es el mismo.

|4x + 1| > 0 ⇒ 4x + 1 > 0 o -(4x + 1) > 0

… ⇒ 4x + 1 > 0 o -4x - 1 > 0

… ⇒ 4x > -1 o -4x > 1

… ⇒ x > -1/4 o x < -1/4

⇒ x ≠ -1/4

|x - 1| < 5 ⇒ x - 1 < 5 o -(x - 1) < 5

… ⇒ x - 1 < 5 o -x + 1 < 5

… ⇒ x < 6 o -x < 4

… ⇒ x < 6 o x > -4

⇒ -4 < x < 6

7 0

2 years ago