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3 years ago

PLEASE HELP!! BRAINLIEST ANSWER AND FASTEST ANSWER GETS TON OF FREE POINTS...MATH WHIZZES, HELP ME OUT. FORGOT HOW TO DO THIS.

Mathematics

1 answer:

Julli [10]3 years ago

5 0

Inscribed angles equal half of their arc length. So <D+ <B = 180
substitute: x+24+x+10 = 180
solve: 2x + 34 = 180
2x= 146
x= 73
NOw find the measure of angle A:
<a = x + 15
<a = 73 + 15
<a = 88
Now to find <C
<C = 180- 88
<C = 92
Hope this helps!

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2 Barbie is thinking of a number.

arlik [135]

Answer:

I think -4

Step-by-step explanation:

20- 1/3of 72

I hope I am right and it helps you!

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4 years ago

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Bryan owns a pet store. He puts one parrot in per cage, he has one parrot extra. If he puts two parrots per cage, he has a cage

Ilya [14]

Answer:

4 parrots

3 cages

Step-by-step explanation:

Bryan has one more parrot than cages and one more cage than half the number of parrots. The following system of equations can be modeled for the number of parrots (P) and cages (C):

$P=C+1\\C-1=\frac{P}{2}$

Solving the linear system:

$P-1-1=\frac{P}{2}\\0.5P = 2\\P=4\\C=P-1=4-1=3$

Bryan has 4 parrots and 3 cages.

6 0

3 years ago

"solve for X" any help would be appreciated thankss

VARVARA [1.3K]

Answer: x = 5

Step-by-step explanation:

8 0

3 years ago

Two mechanics worked on a car. the first mechanic worked for 15 hours, and the second mechanic worked for 5 hours. together they

Stells [14]

Let
x----------> the first mechanic rates
y----------> the second mechanic rates

we know that
x+y=165
15x+5y=1775

using a graph tool
see the attached figure

the solution is
x=95
y=70

the answer is

the rates of the first mechanic is 95
the rates of the second mechanic is 70

4 0

3 years ago

.. Which of the following are the coordinates of the vertices of the following square with sides of length a?

atroni [7]

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Step-by-step explanation:

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

To find the sides of a square, let us use the distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

Thus, the square with vertices $O(0,0), S(a,0), T(a,a), W(0,a)$ has sides of length a.

Thus, the correct answers are option a and option d.

8 0

3 years ago