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liberstina [14]
2 years ago
14

Please help with this

Mathematics
1 answer:
PSYCHO15rus [73]2 years ago
3 0

Answer:

B. m \angle1 + m \angle3

D. 180\degree - m \angle6

Step-by-step explanation:

By exterior angle theorem:

m \angle4 = m \angle1 + m \angle3

m \angle4 = 180\degree - m \angle6

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0.00045−2.5⋅10 −5 = what is the answer i been up all day
Mama L [17]
If I’m getting this right the equation is: 0.00045 - 2.5 x 10 - 5? The dot in between 2.5 and 10 indicates a multiplication sign

So...

0.00045 - (2.5 x 10) - 5
0.00045 - 25 - 5
= -29.9999 -> -30
8 0
3 years ago
I WILL GIVE BRAINLIEST, PLZ HELP!<br><br> given g(x) = - x-4, solve for x when g(x) =-4
yaroslaw [1]

Answer:

g(-4) = 0

Step-by-step explanation:

g(-4) = -(-4)-4

g(-4) = 0

5 0
3 years ago
P = -2p - 15<br><br> I need help plz. Thanks!
mihalych1998 [28]

Answer: p=−5

Step-by-step explanation:

Step 1: Add 2p to both sides.

p+2p=−2p−15+2p

3p=−15

Step 2: Divide both sides by 3.

3p/3 = −15/3

p=−5

4 0
3 years ago
Read 2 more answers
What is (are) the solution(s) to the equation 23x−3=−13x−6?
solniwko [45]
From Question ,
23x-3 = 13x-6
Or, 23x-13x = -6+3
Or, 10x = -3
Or, x = -3/10
Therefore x = -3/10 ans

Hope this helps !

Please mark it as the brainliest
6 0
3 years ago
Peter's farm has 160 meters of fencing, and he wants to fence a rectangular field thatborders a straight river. He needs no fenc
Likurg_2 [28]

ANSWER

3200 m²

EXPLANATION

Peter wants to build a fence around a rectangular field, except for one side because the river is there,

So, the total perimeter of the fence is,

P=W+W+L=2W+L

And this is equal to 160 m. Solving for L,

160=2W+L\text{ }\Rightarrow\text{ }L=160-2W

The area of this field is,

A=WL

Replace L with the expression we found from the perimeter,

A=W(160-2W)=160W-2W^2

The area is given by a quadratic function whose leading coefficient is negative, which means that the graph is a downward parabola and, therefore, the vertex is the maximum value of the area.

The x-coordinate of the vertex is given by,

f(x)=ax^2+bx+c\text{ }\Rightarrow\text{ }x_{vertex}=\frac{-b}{2a}

In this case, x is W, a = -2, and b = 160, so the width of the field for the maximum area is,

W_{vertex}=\frac{-160}{2(-2)}=\frac{160}{4}=40m

And the length when W = 40 is,

L=160-2W=160-2\cdot40=160-80=80m

And the area is,

A=WL=40m\cdot80m=3200m^2

Hence, the largest area of Peter's farm that can be fenced is 3200 square meters.

5 0
1 year ago
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