All categories

2 years ago

Given the central angle, name the arc formed. ∠GQJ arc circle 13 A. FHˆ B. IJˆ C. GJˆ D. GQJˆ

Mathematics

1 answer:

Vaselesa [24]2 years ago

8 0

The name of the arc formed by the central angle ∠GQJ is (c) GJ

The central angle is given as:

∠GQJ

The above means that the center of the circle is located at point Q.

When the point Q is removed from the name of the central angle, we have points G and J

These represent the endpoints of the arc.

Hence, the name of the arc formed by the central angle ∠GQJ is (c) GJ

You might be interested in

If your insurance plan is $15,000 and you pay for 20%, how much do you

podryga [215]

Answer: 250

Step-by-step explanation:

20% of 15,000 is 3,000

3,000/12=250

7 0

3 years ago

Read 2 more answers

Evaluate 7 + (-3X2) for x = 0.

belka [17]

Use pemdas for these types of questions. -3x2= 6 7+6 = 13 x=13

3 0

4 years ago

Help please this i will mark brainliest please

solniwko [45]

Step-by-step explanation:

please complete your question

6 0

3 years ago

Hey plz help meeeeeeeeeeeeeeeeeeeee plzzzzzzzzzzzzz

Alex777 [14]

3400$-25%=2550$(remainder)

2550$*30%=765$(spent on transportation)

6 0

3 years ago

What is the quotient?

exis [7]

The quotient is $\frac{3(y-5)}{2}$

Explanation:

The expression is $\frac{2 y^{2}-6 y-20}{4 y+12} \div \frac{y^{2}+5 y+6}{3 y^{2}+18 y+27}$

Now, reciprocal the term $\frac{y^{2}+5 y+6}{3 y^{2}+18 y+27}$ and convert ÷ to ×

$\frac{2 y^{2}-6 y-20}{4 y+12} \times \frac{3 y^{2}+18 y+27}{y^{2}+5 y+6}$

Taking the common terms out, we get,

$\frac{2\left(y^{2}-3 y-10\right)}{4(y+3)} \times \frac{3\left(y^{2}+6 y+9\right)}{y^{2}+5 y+6}$

Factorizing each numerator and denominator, we get,

$\frac{2(y-5)(y+2)}{4(y+3)} \times \frac{3(y+3)(y+3)}{(y+3)(y+2)}$

Multiplying the terms, we have,

$\frac{6(y-5)(y+2)(y+3)^{2}}{4(y+3)^{2}(y+2)}$

Cancelling the common factors, we get,

$\frac{3(y-5)}{2}$

Thus, the quotient is $\frac{3(y-5)}{2}$

8 0

3 years ago