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oksano4ka [1.4K]
2 years ago
9

Leyla drops a penny from a height of 150m.

Mathematics
1 answer:
dusya [7]2 years ago
5 0
Asked and answered elsewhere.
brainly.com/question/9071599
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Someone plz help me out i need to pass
Anni [7]

Answer:

it takes 24h

Step-by-step explanation:

3 0
3 years ago
Which of the following points lie in the solution set to the following system of inequalities? (1 point)
sineoko [7]
(1,-10) 
the 10 must be negative

reason why the others dont work:
(1,10) too positive, the other two start negative, no way is it going to be way up here
(10,1) still too high
(-1,10) then again, too high

4 0
3 years ago
Read 2 more answers
State the domain restriction(s) in interval notation of \displaystyle f\left(g\left(x\right)\right)f(g(x)) given: \displaystyle
DENIUS [597]

Answer:

The interval notation for the domain is [\frac{23}{3},\infty  ].

Step-by-step explanation:

Consider the provided information.

It is given that \:f\left(x\right)=\sqrt{3x-2},\:\text{ and }\:g\left(x\right)=x-7

We need to find the value of f\left(g\left(x\right)\right).

Put the value of g(x) in  f\left(g\left(x\right)\right).

f\left(g\left(x\right)\right)=f(x-7)  ....(1)

Now, put x=x-7 in \:f\left(x\right)=\sqrt{3x-2}

\:f\left(x-7\right)= \sqrt{3(x-7)-2}

\:f\left(x-7\right)= \sqrt{3x-21-2}

\:f\left(x-7\right)= \sqrt{3x-23}

From equation 1.

f\left(g\left(x\right)\right)=\:f\left(x-7\right)= \sqrt{3x-23}

The domain of the function is the set of input values for which a function is defined.

Here, the value of 3x-23 should be greater or equal to 0 as the square root of a negative number is not real.

Domain= 3x-23\geq0

x\geq\frac{23}{3}

The value of x is all real number greater than \frac{23}{3}.

Hence, the interval notation for the domain is [\frac{23}{3},\infty  ].

7 0
2 years ago
What value of c makes x^2+6+c a perfect square trinomial ?
Citrus2011 [14]

According to this, then:

6x = 2*sqrt(c)*x

6 = 2*sqrt(c)

3 = sqrt(c)

c = 3^2 = 9

Hope this helps!

6 0
3 years ago
Read 2 more answers
A watercolor painting is 24 inches long by 11 inches wide. Ramon makes a border around the watercolor painting by making a mat t
sergeinik [125]

Given: It is given that the length of the painting is 24 inches and the width is  11 inches.

To find: Area of the mat

Solution:

The watercolor painting is 24 inches long by 11 inches wide.

So, the area of the painting is:

\text{Area of painting}=l\times b

\text{Area of painting}=24\times 11

\text{Area of painting}=264 \text{ in}^2

The length of painting with mat is = 24 in + 3 in + 3 in = 30 in

The width of painting with mat = 11 in + 3 in + 3 in = 17 in

\text{Area of painting with mat}=30\times 17

\text{Area of painting with mat}=510 \text{ in}^2

Now to calculate the area of mat subtracts the area of painting from the area of the mat.

\text{Area of mat}=\text{Area of painting with mat}-\text{Area of painting}\\

\text{Area of mat}=510-264

\text{Area of mat}=246 \text{ in}^2

Hence, the area of the mat is 246 in².

6 0
2 years ago
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