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2 years ago

Can someone please give me the (Answers) to this? ... please ...

Mathematics

1 answer:

Igoryamba2 years ago

5 0

Step-by-step explanation:

I have attached a photo. That might help you out. If still anything is unclear, feel free to let me know in the comments section.

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The length of the rectangle is 3 cm and the width is 5 cm what will the area be?

WINSTONCH [101]

Answer:

15 cm²

Step-by-step explanation:

3(5)=15 cm²

3 0

3 years ago

The model below represents the equation 3x+2=2x-5,

Leno4ka [110]

Step-by-step explanation:

Hey there!

The equation is;

$3x + 2 = 2x - 5$

Shifting "2x" to left side and "2" to right side.

$3x - 2x = - 5 - 2$

Simplify them to get answer.

$x = - 7$

Therefore the value of x is -7.

Check;

3 × -7 + 2 = 2 × -7 -5

-19 = -19 ( True ).

So, your answer is option B.

Hope it helps..

7 0

3 years ago

What do you call identical twin sisters when both are ice skating champions

BigorU [14]

The answer for your question is Ice queen clones.

3 0

3 years ago

Read 2 more answers

Write and solve an inequality.

TEA [102]

Answer:

Step-by-step explanation:

4 0

2 years ago

Find the exponential function that passes through the points (2,80) and (5,5120)

juin [17]

Answer:

y = 5·4^x

Step-by-step explanation:

If you have two points, (x1, y1) and (x2, y2), whose relationship can be described by the exponential function ...

y = a·b^x

you can find the values of 'a' and 'b' as follows.

Substitute the given points:

y1 = a·b^(x1)

y2 = a·b^(x1)

Divide the second equation by the first:

y2/y1 = ((ab^(x2))/(ab^(x1)) = b^(x2 -x1)

Take the inverse power (root):

(y2/y1)^(1/(x2 -x1) = b

Use this value of 'b' to find 'a'. Here, we have solved the first equation for 'a'.

a = y1/(b^(x1))

In summary:

b = (y2/y1)^(1/(x2 -x1))
a = y1·b^(-x1)

For the problem at hand, (x1, y1) = (2, 80) and (x2, y2) = (5, 5120).

b = (5120/80)^(1/(5-2)) = ∛64 = 4

a = 80·4^(-2) = 80/16 = 5

The exponential function is ...

y = 5·4^x

3 0

3 years ago

￼Can someone please give me the (Answers) to this? ... please ...

Can someone please give me the (Answers) to this? ... please ...