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3 years ago

if an angle is acute, then the measure of its complement must be greater than the measure of its supplement? a.true b.false

1 answer:

iogann1982 [59]3 years ago

6 0

The answer is a because the compliment is the other part of the 180 degree half circle, and if it's acute it has to be smaller than its compliment

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Help plzzz..............

Gennadij [26K]

A
area = 97*79=7663 ft^2
perimeter = 2(97)+2(79) = 194+158 = 352 ft

b perimeter

c area

8 0

3 years ago

If m\angle A=(6x+10)^{\circ}∠A=(6x+10) ∘ and m\angle B=(4x+30)^{\circ}∠B=(4x+30) ∘ , then find the measure of \angle B∠B.

netineya [11]

A=6 × + 10=88555%__/455

6 0

2 years ago

What percent is equivalent to 7/20

expeople1 [14]

35%
divide whatever number you get move the decimal two spaces to the right and that is the percent.

4 0

3 years ago

10 less than 5 times a number is less than 6 times the number decreased by eight

-BARSIC- [3]

Answer:

The Answer is: the Number is -2

Step-by-step explanation:

Let n = the number.

Five times the number minus 10 is less than 6 times the number minus eight:

5n - 10 = 6n - 8

-1n = -8 + 10

-n = 2

n = -2, the value of the number is -2.

Proof:

5(-2) - 10 = 6(-2) - 8

-10 - 10 = -12 - 8

-20 = -20

Hope this helps! Have an Awesome Day!! :-)

8 0

3 years ago

Write the equation in standard form for the circle with center (0, – 10) and radius 6.

Artemon [7]

Answer:

Equation of circle in standard form with centre (0, – 10) and radius 6. $\mathbf{x^2+(y+10)^2=36}$

Step-by-step explanation:

We need to write equation of circle in standard form with centre (0, – 10) and radius 6.

The equation of circle is: $(x-h)^2+(y-k)^2=r^2$

where (h,k) is centre of circle and r is radius

We are given centre (0,-10) so, h=0, k=-10 and radius r = 6

Putting values and finding equation:

$(x-h)^2+(y-k)^2=r^2\\(x-0)^2+(y-(-10))^2=(6)^2\\x^2+(y+10)^2=36$

So, equation of circle in standard form with centre (0, – 10) and radius 6. $\mathbf{x^2+(y+10)^2=36}$

5 0

2 years ago