All categories

3 years ago

Triangle ABC is equilateral. Find the value of x.

Mathematics

1 answer:

AfilCa [17]3 years ago

7 0

Worked solution on photo

You might be interested in

A store owner bought a painting for $120. She is selling it for $190. What is the percent markup on the painting

stiks02 [169]

Answer:

The markup percent on the painting is 58.3%

Step-by-step explanation:

Here, The Cost price of the painting = $120

The selling price of the painting = $190

Hence, the total profit on the painting = Cost Price - Selling price

or Profit = $190 - $120 = $70

Now, Markup % = (Gross Profit / cost) x 100

⇒Markup % = $\frac{70}{120} \times 100 = 58.3$

Hence, the markup percent on the painting is 58.3%

7 0

3 years ago

What is the surface area of a cone

Kisachek [45]

I do not know I’m sorry

8 0

4 years ago

Brainliest to whoever is correct 30pts...!!!!!!!!!

LiRa [457]

Angle between 16 and 18 = 90 + 55 = 145°

$d= \sqrt{16^2+18^2-2*16*18*cos(145^o)}= \\ \\ \sqrt{256+324+471.83}= \sqrt{1051.83} \approx 32$

5 0

3 years ago

On a trip, a car traveled 80 miles in an hour and a half, then was stopped in traffic for 30 minutes, then traveled 100 miles du

Wewaii [24]

Answer:

45 mph

Step-by-step explanation:

6 0

3 years ago

A Mexican restaurant sells quesadillas in two sizes: a "large" 10 inch-round quesadilla and a "small" 5 inch-

elena-14-01-66 [18.8K]

Answer:

Half of the 10-inch quesadilla is greater than the entire 5-inch quesadilla

Step-by-step explanation:

we know that

The area of a circle is equal to

$A=\pi r^2$

where

r is the radius

step 1

Find the area of the 10 inch-round quesadilla

we have

$D=10\ in$

$r=10/2=5\ in$

substitute

$A=\pi (5)^2$

$A=25\pi\ in^2$

step 2

Find the area of the 5 inch-round quesadilla

we have

$D=5\ in$

$r=5/2=2.5\ in$

substitute

$A=\pi (2.5)^2$

$A=6.25\pi\ in^2$

step 3

Which is larger, half of the 10-inch quesadilla or the entire 5-inch quesadilla?

Compare

half of the 10-inch quesadilla is equal to ----> $\frac{1}{2}(25\pi)=12.5\pi\ in^2$

the entire 5-inch quesadilla ----> $6.25\pi\ in^2$

therefore

Half of the 10-inch quesadilla is greater than the entire 5-inch quesadilla

3 0

3 years ago