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Ede4ka [16]
3 years ago
8

PLEASE HELP ME!!!

Mathematics
1 answer:
melisa1 [442]3 years ago
7 0

Answer:

1.2

Step-by-step explanation:

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Which term describes the figure? Choose 1 answer: Parallel lines Perpendicular lines None of the above​
liubo4ka [24]

Answer:

perpendicular lines

Step-by-step explanation:

because they intersect, they do not go even next to each other forever.

7 0
3 years ago
In the triangle below, determine the value of a.
dusya [7]

Answer:

6.1

Step-by-step explanation:

Taking cos function,

Cos C=BC/AC

Cos 35°=5/a

a=5/cos 35°

a=6.1

5 0
4 years ago
Which of the following is not equal to sin⁡(-230°)?
xenn [34]

From trigonometry we know that:

if sin(\theta)=sin(\alpha)

then, \theta=n\pi+(-1)^n\alpha (where n is an integer)

This can be rewritten in degrees as:

\theta=n(180^{\circ})+(-1)^n\alpha.............(Equation 1)

Now, in our case, \alpha=-230^{\circ}

Therefore, (Equation 1) can be written as:

\theta=n(180^{\circ})+(-1)^n(-230^{\circ})..........(Equation 2)

Now, to find the correct options all that we have to do is replace n by relevant integers and find the values of \theta that match.

For n=2, (Equation 2) gives us: \theta=2\times 180^{\circ}+(-1)^2(-230^{\circ})=360^{\circ}-230^{\circ}=130^{\circ}.

Thus, sin(230^{\circ})=sin(130^{\circ})

Now, we know that: -sin(-50^{\circ})=sin(50^{\circ})

Let n=-1, then:

\theta=(-1)\times 180^{\circ}+(-1)^{-1}(-230^{\circ})=-180^{\circ}+230^{\circ}=50^{\circ}

Thus, sin(-230^{\circ})=-sin(-50^{\circ})

Likewise, sin(-230^{\circ})=sin(50^{\circ})

Only the last option sin(-50^{\circ}) will never match sin(-230^{\circ}) because no integral value of n will ever give \theta=-50^{\circ}

Thus the last option is the correct option.

3 0
3 years ago
Read 2 more answers
Let f(x)=5x+5 find f(-1)<br><br>I need your help with this. please help me​
Alisiya [41]

f(x) = 5x+5\\\\f(-1) = 5(-1) +5  = -5+5=0

5 0
3 years ago
Cristine's middle school has a total of 900 students and 45 teachers. The local high school has 110 teachers and a student-teach
RUDIKE [14]
Well what I would do first is divide the 900 student by the 445 teacher to get 20 students per teacher. Now I would take that 20 and multiply it by 110 teacher to get 2200 Students.

Answer= C) 2200 Students
4 0
3 years ago
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