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Oliga [24]
2 years ago
14

Solve this: -4<3x-4<_ 11 A. 5 B. 0 C. 0

Mathematics
1 answer:
statuscvo [17]2 years ago
4 0
Your answer most likely will be A
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2/3 as a real number
BaLLatris [955]

Answer:

Yes

Step-by-step explanation:

I don't know the question but 2/3 is a real number.

8 0
3 years ago
CAN SOMEONE PLEASE HELP ME WITH MY GEOMETRY HOMEWORK PLEASE?
Kay [80]

Answer:

Option A

Step-by-step explanation:

The complete question is shown in the attachment.

The sum of angles in a triangle is 180 degrees.

This implies that:

m\angle A+85+53=180

m<A+138=180

m<A=180-138=42

Now we use the sine rule to find AC=b

\frac{b}{\sin 85}=\frac{85}{\sin 42}

This implies that:

b=\frac{85}{\sin 42}*\sin 85

b=126.5

5 0
3 years ago
What is an equation of the line that passes through the point (-4, -6) and is
const2013 [10]
Answer:y=2/3x-10/3
- 4/-6 is 4/6 so m is 4/6 then fill in the x and y variables in y=Mx+b solve for b then start plotting the equation of the line to check and you get your answer

I think this is the answer you might need to check again with more people
8 0
2 years ago
Read 2 more answers
Help I don't know what to choose
Lunna [17]

Answer:

Step-by-step explanation:

We know a triangle adds up to 180 so:

75+36+ ? = 180

111+ ? =180

? = 69

To find out x, we know that a straight line adds up to 180 so again:

x +69 = 180

x = 111

8 0
2 years ago
Read 2 more answers
Use lagrange multiplier techniques to find the local extreme values of f(x, y) = x2 − y2 − 2 subject to the constraint x2 + y2 =
dexar [7]
Given f(x,\ y)=x^2-y^2-2 subject to the constraint x^2+y^2=16

Let g(x,\ y)=x^2+y^2.

The gradient vectors of f and g are:

\nabla f(x,\ y)=\langle2x,-2y\rangle and \nabla g(x,\ y)=\langle2x,2y\rangle

By Lagrange's theorem, there is a number \lambda, such that

\langle2x,-2y\rangle=\lambda\langle2x,2y\rangle=\langle2\lambda x,2\lambda y\rangle

\lambda=\pm1

It can be seen that f(x,\ y)=x^2-y^2-2 has local extreme values at the given region.
6 0
3 years ago
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