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Tatiana [17]
3 years ago
7

PLEASE HELP I CAN'T FIGURE IT OUT

Mathematics
1 answer:
Vlada [557]3 years ago
7 0
It’s very easy lol it’s B
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When asked to find a number one-tenth as large as another number, what operation would you use? What about when asked to find a
kherson [118]

Apply division by 10 when one tenth of a number is required and apply multiplication by 10 when 10 times of a number is required.

<u>Solution:</u>

Need to determine what operation is required to get one-tenth of a number and 10 times of a number

To get one tenth of a number, divide the number by 10.

For example to get one – tenth of 100, divide it by 10, we get 10 as a result.

\frac{1}{10} \times 100 = 10

To get ten times of a number, multiply the number by 10

For example 10 times of 10 = 10 x 10 = 100

Hence apply division by 10 when one tenth of a number is required and apply multiplication by 10 when 10 times of a number is required.

6 0
3 years ago
The diameter of a circle is 5 ft. Find the circumference to the nearest tenth(urgent)
defon

Answer:

15.71

Step-by-step explanation:

To find the radius, half the diameter because r x 2 = diameter. 5/2 = 2.5

Put in circumference formula C=2πr

c=2 x 3.14 x 2.5

c= 2 x 7.85

c=15.7

Put the value into the circle circumference formula C=2πr

5 0
3 years ago
COULD SOMEONE PLS HELP ME WITH MY LATEST 2 QUESTIONS?
OlgaM077 [116]
I can try how do I help you ?
4 0
3 years ago
The domain of ​(f​g)(x) consists of the numbers x that are in the domains of both f and g.
Dovator [93]

The statement "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g" is FALSE.

Domain is the values of x in the function represented by y=f(x), for which y exists.

THe given statement is "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g".

Now we assume the g(x)=x+2 and f(x)=\frac{1}{x-6}

So here since g(x) is a polynomial function so it exists for all real x.

f(x)=\frac{1}{x-6}<em>  </em>does not exists when x=6, so the domain of f(x) is given by all real x except 6.

Now,

(fg)(x)=f(g(x))=f(x+2)=\frac{1}{(x+2)-6}=\frac{1}{x-4}

So now (fg)(x) does not exists when x=4, the domain of (fg)(x) consists of all real value of x except 4.

But domain of both f(x) and g(x) consists of the value x=4.

Hence the statement is not TRUE universarily.

Thus the given statement about the composition of function is FALSE.

Learn more about Domain here -

brainly.com/question/2264373

#SPJ10

3 0
1 year ago
Solve for y.<br><br> −140=18+4(5y−2)
vagabundo [1.1K]

For this case we have the following equation:

-140 = 18 + 4 (5y-2)

From here, we must clear the value of y.

For this, we follow the following steps:

1) Distributive property:

-140 = 18 + 20y-8

2) Variable terms on one side of the equation and constant terms on the other side of the equation:

20y = -140 - 18 + 8

3) Algebraically add the constant terms:

20y = -150

4) Pass the number 20 to divide:

y = -\frac{150}{20}\\y = -7.5

Answer:

The value of y is given by:

y = -7.5

3 0
3 years ago
Read 2 more answers
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