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2 years ago

Help me plz and show work <3

Mathematics

1 answer:

icang [17]2 years ago

4 0

Answer:

y = (5-2x)/7 = 5/7 - 2x/7. option C

Step-by-step explanation:

here's your solution

=> 7y + 2x = 5

=> 7y = 5 - 2x

=> y = (5-2x)/7

$\huge\mathfrak\red{hope \: it \: helps...}$

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PLEASE HELP I WILL GIVE BRAINLIEST

OLga [1]

Answer:

$67

Step-by-step explanation:

So 5x12+7=C

PEDMAS so:

5x12 = 60

60+7 = 67

67 = C

8 0

2 years ago

Please help, I would appreciate it a lot

AleksAgata [21]

Answer:

Sry, Can't help.

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2 years ago

Two families go skiing on a Saturday. One family purchases three adult lift tickets and four lift tickets for $188. Another fami

Mnenie [13.5K]

Answer:using elimination to solve for x

9x +6y = 3*134 | Multiplying 1st equation by 3

-4x -6y = -2*146 | Multiplying 2nd equation by -2

5x = 3*134 - 2*146

5x = $110

x = $22, the cost of the youth ticket

Step-by-step explanation:This question Sets Up two equations with 2 unknowns

Let x and y represent the cost of the youth and adult ticket respectively

Question states***

3x + 2y = $134

2x + 3y = $146

7 0

2 years ago

Find the graph of the inequality y+2>-3(x+1)

Serhud [2]

Answer:

Below.

Step-by-step explanation:

y + 2 > -3(x + 1)

y + 2 > -3x - 3

y > -3x - 5

The graph will be the graph of the line y = -3x - 5 (dotted line), with the area above the line shaded.

3 0

2 years ago

Is domain of f(x,y) = 1 + (4 -y^2)^1/2 open, closed or neither<br><br> is it bounded or unbounded?

omeli [17]

Answer:

Closed and Bounded.

Step-by-step explanation:

Hi there!

1) Let's start by finding the values in which the function is defined. Remember that this can be rewritten:

$f(x,y) = 1 + (4 -y^2)^{\frac{1}{2}} \:id\:est\:f(x,y)=1+\sqrt{(4 -y^2)}$

Since every quadratic root are defined for values $\geq$ 0 then, this help us to understand that we need calculate what interval this Domain is:

$4-y^{2} \geq 0\\4-y^{2}-4\geq -4\\-y^{2}\geq-4 \\y^{2}\leq4\\-2\leq y\leq 2$

$D=[-2,2]$

2) Graphically speaking, the domain is closed. For the values -2 and 2 are included, and bounded.

Bounded functions have all of their points contained by some circle origin centered. Check it out below.

6 0

3 years ago