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fredd [130]
4 years ago
15

Help? With all? Plzzz

Mathematics
1 answer:
Margaret [11]4 years ago
7 0
Cuz if so, then you just move the whole x term to the right side and divide all the terms on the right side by the coefficient of y
for ex...
3) 4x+4y=16
4y=16-4x (or -4x+16)
y=4-x or -x+4
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Nitella [24]

Answer:

no.

Step-by-step explanation:

the 4 on the sides are half the area of the ones of the top and bottom because if you take two small ones they will fit with one of the big ones.

sorry if I'm wrong.

5 0
3 years ago
Drag an answer to each box to complete this paragraph proof.
sveticcg [70]

1.) m∠P+m∠Q+m∠R=180

2.) x=10


4 0
3 years ago
Read 2 more answers
-6+x/x^4 limit at x →0
Jlenok [28]

Answer:

not existing

Step-by-step explanation:

\frac{ - 6 + x}{ {x}^{4} }

lim x goes to 0 is

\frac{ - 6 + 0}{ {0}^{4} }  =   \frac{ - 6}{0}

so not exist

I'm not sure if u mean that or this

7 0
3 years ago
Does a equilateral triangle have rotational symmetry if so identify the angle of rotation
nydimaria [60]
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5 0
4 years ago
Which two points are on the graph of M(a) = - 3/4a-12?
sertanlavr [38]

Answer:

(-16, 0) and  (0,-12)  are exactly two points on the graph of the given equation.

Step-by-step explanation:

Here, the given expression is : M(a) = -\frac{3}{4} a-12

Here, let M(a)  = b

⇒The equation becomes b = -\frac{3}{4} a-12

Now, check for all the given points for (a,b)

<u>1) FOR (-9,0)</u>

RHS  is  -\frac{3}{4} (-9)-12  =   -5.25 \neq 0

Hence, (-9,0) is NOT on the graph.

<u>2) FOR (-16,0)</u>

Here, LHS = b = 0

andRHS  =   -\frac{3}{4} (-16)-12  = 12-12  = 0

Hence, LHS  =  RHS = 0  So, (-16,0) is  on the graph.

<u>3) FOR (0,12)</u>

Here, LHS = b = 12

RHS  is  -\frac{3}{4} (0)-12  =   -12  \neq 12

Hence, (0,12) is NOT on the graph.

<u>4) FOR (0,-12)</u>

Here, LHS = b = -12

RHS  is  -\frac{3}{4} (0)-12  =   -12

and LHS = RHS = -12

Hence, (0,-12) is  on the graph.

Hence, (-16, 0) and  (0,-12)  are exactly two points on the graph of the given equation.

6 0
3 years ago
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