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JulijaS [17]
3 years ago
13

Please help me Fill in the blank with a variable term to make the equation true

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
3 0

Answer:

x = 6

2x+3=15

2x=12

x=6

therefore,

6x+3=15

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Solve for x in the diagram below
BlackZzzverrR [31]

Answer:

40

Step-by-step explanation:

because vertical angles are equal

3×40=120

4 0
3 years ago
-3x - 6+ (-1)<br> help!!!!
guapka [62]

Answer:

-3x = 7 .  

Step-by-step explanation:

-3x - 6 - 1 =

-3x - 7

Hope that helps!

5 0
3 years ago
Read 2 more answers
How do I solve the equation?
trapecia [35]

Answer:

(-3, 5) radius- r= 5

Step-by-step explanation:

x^2+6x+y^2-10y=-9

Rewrite ^2+6x+y^2-10y=-9 in the form of the standard circle equation

(x-(-3))^2+(y-5)^2= 5^2

Therefore the circle properties are:

(a, b)=(-3, 5), r= 5

Download pdf
5 0
2 years ago
Which of the following is the graph of y=sqr root -x-3
Elan Coil [88]

Answer:

The graph in the attached figure

see the explanation

Step-by-step explanation:

we have

y=\sqrt{-x-3}

we know that

The radicand must be greater than or equal to zero

so

(-x-3)\geq 0

solve for x

Adds 3 both sides

-x\geq 0+3

-x\geq 3

Multiply by -1 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

x\leq -3

so

The domain of the function is the interval (-∞,-3]

For x=-3 ---> the value of y=0

The range is the interval {0,∞)

therefore

The graph in the attached figure

3 0
3 years ago
HELP ME QUICK PLEASEE!!
Lena [83]

Answer:

  A) is a function

  B) f(x) is greater

  C) x = 12

Step-by-step explanation:

A) Yes, the relation is a function.

A function has a unique input value for each output value. There are no repeated x-values, so this relation is a function.

__

B) The table says y=6 for x=6.

  f(6) = 2(6)+16 = 28

f(x) has a greater value for x=6.

__

C) Substituting the given information, we have ...

  40 = 2x +16

  24 = 2x

  12 = x

The value of x is 12 when f(x) is 40.

3 0
3 years ago
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