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3 years ago

How do you do the number line part ?

Mathematics

1 answer:

Naily [24]3 years ago

5 0

Answer:

you would mark the different points less than 250 and then you plot the inequality from the previous questions.

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Which of the following is equivalent to –2i(6 – 7i)? (0 – 2i)(6 – 7i) (0 + 2i)(6 – 7i) (–2 + i)(6 – 7i) (6 – 2i) – (7i – 2i)

Mariulka [41]

Answer:

The expression which is equivalent to –2i(6 – 7i) is; (0 – 2i)(6 – 7i)

4 0

2 years ago

What does y=2x+3 equal?

posledela

Y = X times 2 + 3 if that's what you're asking so y = 5x
Explanation:
2+3=5 so 5x bc x isn't a number

4 0

3 years ago

F - 3/4 = 1/2 what is f?

Julli [10]

Answer:

f=5/4

Step-by-step explanation:

f=1/2+3/4

f=(1/2)×2+3/4

f=2/4+3/4

f=5/4

3 0

3 years ago

Read 2 more answers

What is -5 1/10 equal to

Delicious77 [7]

-5 and 1/10
try to convert tomixed fraction
1=10/10 in this problem
-5=-50/10
add
-50/10-1/10=-51/10

another form is decimal form
-5 and 1/10
1/10=0.1
-5.1 is another answer

3 0

3 years ago

Read 2 more answers

Find the exact value of each trigonometric function for the given angle θ.

Kay [80]

Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

Therefore, $\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.$

8 0

3 years ago