All categories

3 years ago

Please help me!! I am struggling... I will not accept nonsense answers!

Mathematics

1 answer:

telo118 [61]3 years ago

6 0

Answer:

y = 110°

Step-by-step explanation:

The inscribed angle CHF is half the measure of its intercepted arc CDF

The 3 arcs in the circle = 360°, thus

arc CDF = 360° - 160° - 60° = 140°, so

∠ CHF = 0.5 × 140° = 70°

∠ CHF and ∠ y are adjacent angles and supplementary, thus

y = 180° - 70° = 110°

You might be interested in

Simplify

Goryan [66]

Answer:

310

Step-by-step explanation:

Use the Order of Operations method. (P.E.M.D.A.S.)

Our expression: 3+2⋅4+62⋅5−1

Multiply 62 and 5, and 2 and 4:

3+8+310-1

Add and Subtract from left to right:

11+310-1

311-1

310

6 0

2 years ago

How do u solve 63=-9d?

Len [333]

-9d = 63
you divide both side by -9

sooo...

d = -7

7 0

4 years ago

-0.38 written as a fraction the 8 is a repeating decimal

Simora [160]

Answer:

$\frac{-7}{18}$

Step-by-step explanation:

The line in between the numerator and denominator serves as a division sign so, -7 divided by 18 equals -0.38 with the 8 repeating

8 0

3 years ago

I don’t understand how this works? Please help, thanks.

Dima020 [189]

Answer:

(B) Is the answer here

Step-by-step explanation:

Well a function’s domain can’t repeat no matter what.

Domain is X.

So the first one doesnt work since it repeats to every single digit that exists.

The second one works becuase the domain aren’t repeated and the range being repeated (or known as y) is but that deosn’t matter.

6 0

3 years ago

Read 2 more answers

Find the equation for the plane that contains the line x=−1+3t , y=1+2t, z=2+4t and is perpendicular to the plane containing the

Ivan

Let $L$ be the line given by the vector equation

$(-1,1,2) + \lambda(3,2,4) \ , \lambda \in \mathbb{R}$ .

First, we use the director vectors of the lines L1 and L2 to get the

vector equation of the plane containing them, which we denote by $\Pi_1$ . This is,

$\\\\\Pi_1 : (1,-1,5) + \alpha (2,-1,6) + \beta (1,1,-3) \ , \alpha, \beta \in \mathbb{R}\\\\\\$

We observe that $\vec{N} = (2,-1,6)\times(1,1,-3) = (-3,12,3) \ne (0,0,0)$ . Therefore, the vector equation of $\Pi_1$ defines a plane and $\vec{N}$ is a normal vector to $\Pi_1.$

Finally, the vector equation for the wanted plane, which we denote by $\Pi$ , is

$\Pi : (-1,1,2) + r(3,2,4) + s(-3,12,3), r,s \in \mathbb{R} \ .$

Thus, if $s = 0$ , then $L \subset \Pi$ and since $\vec{N}$ is parallel to $\Pi$ , then it is perpendicular to $\Pi_1$ .

8 0

3 years ago